Elements  of  Mechanism 


PETER  SGHWAMB,  S.B. 

Professor  of  Machine  Design  (Rmred),  Massachusetts  Institute  of  Technology 

ALLYNE  L.  MERRILL,  S.B. 

Professor  of  Mechanism,  Massachusetts  Institute  of  Technology 

WALTER  H.  JAMES,  S.B. 

Assistant  Professor  of  Mechanical  Engineering  Drawing, 
Massachusetts  Institute  of  Technology 


THIRD  EDITION,  REWRITTEN,  ENLARGED,  AND  RESET 
TOTAL   ISSUE,    THIRTY   THOUSAND 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC. 

LONDON:   CHAPMAN  &  HALL,   LIMITED 
1921 


Engineering 
Library 


COPYRIGHT,  1904,  1905, 

BY 

PETER  SCHWAMB 

AND 

ALLYNE  L.  MERRILL 


COPYRIGHT,  1921, 

BY 

PETER  SCHWAMB 

ALLYNE  L.  MERRILL 

WALTER  H.  JAMES 


ATLANTIC  PRINTING  COMPANY 
CAMBRIDGE,    MASS.,     U.    8.    A. 


PREFACE  TO  THE  THIRD  EDITION 

This  revised  edition  embodies  many  changes  suggested  by  instructors 
who  have  used  the  second  edition  during  the  last  sixteen  years.  The 
authors  acknowledge  their  indebtedness  to  all  of  these  gentlemen,  and 
especially  to  Prof.  George  W.  Swett,  who  has  read  a  large  part  of  the 
manuscript  and  given  much  valuable  criticism. 

Professor  Walter  H.  James,  in  charge  of  the^instruction  in  Mechanical 
Engineering  Drawing  and  Machine  Drawing  at  the  Massachusetts 
Institute  of  Technology,  and  also  an  Instructor  in  Mechanism  for  many 
years,  has  been  in  charge  of  the  revision,  and  joins  in  the  authorship 
of  the  book. 

PETER  SCHWAMB. 
ALLYNE  L.  MERRILL. 
WALTER  H.  JAMES. 
CAMBRIDGE,  MASS., 
December,  1920 


434J09 


111 


PREFACE  TO   SECOND  EDITION 

The  main  subject-matter  of  this  work  was  written  during  1885  by 
Peter  Schwamb  and  has  been  used  since  then,  in  the  form  of  printed 
notes,  at  the  Massachusetts  Institute  of  Technology,  as  a  basis  for 
instruction  in  mechanism,  being  followed  by  a  study  of  the  mechanism 
of  machine  tools  and  of  cotton  machinery.  The  notes  were  written 
because  a  suitable  text-book  could  not  be  found  which  would  enable 
the  required  instruction  to  be  given  in  the  time  available.  They  have 
accomplished  the  desired  result,  and  numerous  inquiries  have  been 
received  for  copies  from  various  institutions  and  individuals  desiring  to 
use  them  as  text-books.  This  outside  demand,  coupled  with  a  desire 
to  revise  the  notes,  making  such  changes  and  additions  as  experience 
has  proved  advisable,  is  the  reason  for  publishing  at  this  time. 

Very  little  claim  is  made  as  to  originality  of  the 'subject-matter 
which  has  been  so  fully  covered  by  previous  writers.  Such  available 
matter  has  been  used  as  appeared  best  to  accomplish  the  object  desired. 
Claim  for  consideration  rests  largely  on  the  manner  of  presenting  the 
subject,  which  we  have  endeavored  to  make  systematic,  clear,  and 
practical. 

Among  the  works  consulted  and  to  which  we  are  indebted  for  sug- 
gestions and  illustrations  are  the  following:  "Kinematics  of  Machin- 
ery," and  "Der  Konstrukteur,"  by  F.  Reuleaux,  the  former  for  the 
discussion  of  linkages,  and  the  latter  for  various  illustrations  of  mechan- 
isms; " Principles  of  Mechanism,"  by  S.  W.  Robinson,  for  the  discus- 
sion of  non-circular  wheels;  " Kinematics,"  by  C.  W.  MacCord,  for 
the  discussion  of  annular  wheels  and  screw-gearing;  "Machinery  and 
Millwork,"  by  Rankine;  "Elements  of  Mechanism,"  by  T.  M.  Goodeve; 
and  "Elements  of  Machine  Design,"  by  W.  C.  Unwin. 

PETER  SCHWAMB. 

ALLYNE  L.  MERRILL. 
October  20,  1904 


Elements  of  Mechanism 


CHAPTER   I 
INTRODUCTION 

1.  A  Machine  is  a  combination  of  resistant  bodies  so  arranged 
that  by  their  means  the  mechanical  forces  of  nature  can  be  compelled 
to  produce  some  effect  or  work  accompanied  with  certain  determinate 
motions.*     In  general,  it  may  be  properly  said  that  a  machine  is  an 
assemblage  of  parts  interposed  between  the  source  of  power  and  the 
work,  for  the  purpose  of  adapting  the  one  to  the  other.     Each  of  the 
pieces  in  a  machine  either  moves  or  helps  to  guide  some  of  the  other 
pieces  in  their  motion. 

No  machine  can  move  itself,  nor  can  it  create  motive  power;  this 
must  be  derived  from  external  sources,  such  as  the  force  of  gravita- 
tion, the  uncoiling  of  a  spring,  or  the  expansion  of  steam.  As  an 
example  of  a  machine  commonly  met  with,  an  engine  might  be  men- 
tioned. It  is  able  to  do  certain  definite  work,  provided  some  external 
force  shall  act  upon  it,  setting  the  working  parts  in  motion.  It  con- 
sists of  a  fixed  frame,  supporting  the  moving  parts,  some  of  which 
cause  the  rotation  of  the  engine  shaft,  others  move  the  valves  distrib- 
uting the  steam  to  the  cylinder,  and  still  others  operate  the  governor 
which  controls  the  engine.  These  moving  parts  are  so  arranged  that 
they  make  certain  definite  motions  relative  to  each  other  when  an 
external  force,  as  steam  pressure,  is  applied  to  the  piston. 

2.  A  Mechanism  is  a  portion  of  a  machine  where  two  or  more 
pieces  are  combined,  so  that  the  motion  of  the  first  compels  the 
motion  of  the  others,  according  to  a  law  depending  on  the  nature  of 
the  combination.     For  example  the  combination  of  a  crank  and  con- 
necting-rod with  guides  and  frame,  in  a  steam  engine,  serving  to  con- 
vert  reciprocating   into   circular   motion,    would   thus   be   called   a 
mechanism. 

The  term  Elementary  Combination  is  sometimes  used  synonymously 
with  A  Mechanism. 

A  machine  is  made  up  of  a  series  or  train  of  mechanisms. 

*  Reuleaux:  Kinematics  of  Machinery. 
1 


MECHANISM 


3.  The  Science  of  Mechanism  treats  of  the  forms  of  the  pieces 
used  in  the  construction  of  machinery  and  the  laws  which  govern  the 
motions  and  forces  existing  in  or  transmitted  by  them. 

The  operation  of  any  machine  depends  upon  two  things:  first,  the 
transmission  of  certain  forces,  and  second,  the  production  of  deter- 
minate motions.  In  designing,  due  consideration  must  be  given  to 
both  of  these,  so  that  each  part  may  be  adapted  to  bear  the  stresses 
imposed  on  it,  as  well  as  have  the  proper  motion  relative  to  the  other 
parts  of  the  machine.  The  nature  of  the  movements  does  not  depend 
upon  the  strength  or  absolute  dimensions  of  the  moving  parts,  as  can 
be  shown  by  models  whose  dimensions  may  vary  much  from  those 
requisite  for  strength,  and  yet  the  motions  of  the  parts  will  be  the  same 
as  those  of  the  machine.  Therefore,  the  force  and  the  motion  may  be 
considered  separately,  thus  dividing  the  science  of  Mechanism  into 
two  parts,  viz.: 

1°  Pure  Mechanism,  which  treats  of  the  motion  and  forms  of  the 
parts  of  a  machine,  and  the  manner  of  supporting  and  guiding  them, 
independent  of  their  strength. 

2°  Constructive  Mechanism,  which  involves  the  calculation  of  the 
forces  acting  on  different  parts  of  the  machine;  the  selection  of  ma- 
terials as  to  strength  and  durability  in  order  to  withstand  these 
forces,  taking  into  account  the  convenience  for  repairs,  and  facilities 
for  manufacture. 

What  follows,  will,  in  general,  be  confined  to  the  first  part,  pure 
mechanism,  or  what  is  sometimes  called  "  the  geometry  of  machinery  "  ; 
but  in  some  cases  the  forces  in  action  will  be  considered. 

The  definition  of  a  machine  might  be  modified  to  accord  with  the 
above,  as  follows: 

A  Machine  is  an  assemblage  of  moving  parts  so  connected  that 
when  the  first,  or  recipient,  has  a  certain  motion,  the  parts  where  the 
work  is  done,  or  effect  produced,  will  have  certain  other  definite 
motions. 

4.  Driver  and  Follower.     That  piece  of  a  mechanism  which  is 
supposed  to  cause  motion  is  called  the  driver,  and  the  one  whose 
motion  is  effected  is  called  the  follower. 

5.  Frame.     The  frame  of  a  machine  is  a  structure  that  supports 
the  moving  parts  and  regulates  the  path,  or  kind  of  motion,  of  many 
of  them  directly.     In  discussing  the  motions  of  the  moving  parts,  it 
is  convenient  to  refer  them  to  the  frame,  even  though  it  may  have, 
as  in  the  locomotive,  a  motion  of  its  own. 

6.  Modes  of  Transmission.     If  the  action  of  natural  forces  of 
attraction  and  repulsion  is  not  considered,  one  piece  cannot  move 


INTRODUCTION  3 

another,  unless  the  two  are  in  contact  or  are  connected  to  each  other 
by  some  intervening  body  that  is  capable  of  communicating  the 
motion  of  the  one  to  the  other.  In  the  latter  case,  the  motion  of  the 
connector  is  usually  unimportant,  as  the  action  of  the  combination 
as  a  whole  depends  upon  the  relative  motion  of  the  connected  pieces. 
Thus  motion  can  be  transmitted  from  driver  to  follower: 

1°  By  direct  contact. 

2°  By  intermediate  connectors. 

7.  Links  and  Bands.     An  intermediate  connector  can  be  rigid  or 
flexible.     When  rigid  it  is  called  a  link,  and  it  can  either  push  or  pull, 
such  as  the  connecting-rod  of  a  steam  engine.     Pivots  or  other  joints 
are  necessary  to  connect  the  link  to  the  driver  and  follower. 

If  the  connector  is  flexible,  it  is  called  a  band,  which  is  supposed  to 
be  inextensible,  and  only  capable  of  transmitting  a  pull.  A  fluid 
confined  in  a  suitable  receptacle  may  also  serve  as  a  connector,  as  in 
the  hydraulic  press.  The  fluid  might  be  called  a  pressure-organ  in 
distinction  from  the  band,  which  is  a  tension-organ. 

8.  Motion  is  change  of  position.     Motion  and  rest  are  necessarily 
relative  terms  within  the  limits  of  our  knowledge.     We  may  conceive 
a  body  as  fixed  in  space,  but  we  cannot  know  that  there  is  one  so 
fixed.     If  two  bodies,  both  moving  in  space,  remain  in  the  same  posi- 
tion relative  to  each  other,  they  are  said  to  be  at  rest,  one  relatively 
to  the  other;  if  they  do  not,  either  may  be  said  to  be  in  motion  rela- 
tively to  the  other. 

Motion  may  thus  be  either  relative,  or  it  may  be  absolute,  provided 
some  point  is  assumed  as  fixed.  In  what  follows,  the  earth  will  be 
assumed  to  be  at  rest,  and  all  motions  referred  to  it  will  be  considered 
as  absolute. 

9.  Path.     A  point  moving  in  space  describes  a  line  called  its  path, 
which  may  be  rectilinear  or  curvilinear.     The  motion  of  a  body  is 
determined  by  the  paths  of  three  of  its  points  not  on  a  straight  line. 
If  the  motion  is  in  a  plane,  two  points  suffice,  and  if  rectilinear,  one 
point  suffices  to  determine  the  motion. 

10.  Direction.     If  a  point  is  moving  along  a  straight  path  the 
direction  of  its  motion  is  along  the  line  which  constitutes  its  path; 
motion  towards  one  end  of  the  line  being  assumed  as  having  positive 
direction  and  indicated  by  a  +  sign,  the  motion  toward  the  other 
end  would  be  negative  and  indicated  by  a   —  sign.     If  a  point  is 
moving  along  a  curved  path,  the  direction  at  any  instant  is  along  the 
tangent  to  the  curve  and  may  be  indicated  as  positive  or  negative, 
as  in  the  case  of  rectilinear  motion.  * 


4  ELEMENTS  OF  MECHANISM 

11.  Continuous  Motion.     When  a  point  continues  to  move  indefi- 
nitely in  a  given  path  in  the  same  direction,  its  motion  is  said  to  be 
continuous.     In  this  case  the  path  must  return  on  itself,  as  a  circle 
or  other  closed  curve.     A  wheel  turning  on  its  bearings  affords  an 
example  of  this  motion. 

12.  Reciprocating  Motion.     When  a  point  traverses  the  same  path 
and  reverses  its  motion  at  the  ends  of  such  path,  the  motion  is  said 
to  be  reciprocating. 

13.  Vibration  and  Oscillation  are  terms  applied  to  reciprocating 
circular  motion,  as  that  of  a  pendulum. 

14.  Intermittent  Motion.     When  the  motion  of  a  point  is  inter- 
rupted by  periods  of  rest,  its  motion  is  said  to  be  intermittent. 

15.  Revolution  and  Rotation.     A  point  is  said  to  revolve  about  an 
axis  when  it  describes  a  circle  of  which  the  center  is  in,  and  the  plane 
is  perpendicular  to,  that  axis.     When  all  the  points  of  a  body  thus 
move,  the]  body  is  said  to  revolve  about  the  axis.     If  this  axis  passes 
through  the  body,  as  in  the  case  of  a  wheel,  the  word  rotation  is  used 
synonymously  with  revolution.     The  word  turn  is  often  used  synony- 
mously with  revolution  and  rotation.     It  frequently  occurs  that  a  body 
not  only  rotates  about  an  axis  passing  through  itself,  but  also  moves  in 
an  orbit  about  another  axis. 

16.  An  Axis  of  Rotation  or  revolution  is  a  line  whose  direction  is  not 
changed  by  the  rotation ;  a  fixed  axis  is  one  whose  position,  as  well  as 
its  direction,  remains  unchanged. 

17.  A  Plane  of  Rotation  or  revolution  is  a  plane  perpendicular  to 
the  axis  of  rotation  or  revolution. 

18.  Direction  of  Rotation  or  revolution  is  defined  by  giving  the  direc- 
tion of  the  axis  and  stating  whether  the  turning  is  Right-handed 
(clockwise)  or  Left-handed  (anti-clockwise). 

19.  Cycle  of  Motions.     When  a  mechanism  is  set  in  motion  and  its 
parts  go  through  a  series  of  movements  which  are  repeated  over  and 
over,  the  relations  between  and  order  of  the  different  divisions  of  the 
series  being  the  same  for  each  repetition,  one  of  these  series  is  called 
a  cycle  of  motions.     For  example,  one  revolution  of  the  crank  of  a 
steam  engine  causes  a  series  of  different  positions  of  the  piston-rod, 
and  this  series  of  positions  is  repeated  over  and  over  for  each  revolu- 
tion of  the  crank. 

20.  Period  of  motion  is  the  time  occupied  in  completing  one  cycle. 

21.  Linear  Speed  is  the  rate  of  motion  of  a  point  along  its  path,  or 
the  rate  at  which 'a  point  is  approaching  or  receding  from  another 
point  in  its  path.     If  the  point  to  which  the  motion  of  the  moving 
point  is  referred,  is  fixed,  the  speed  is  the  absolute  speed  of  the  point. 


INTRODUCTION  5 

If  the  reference  point  is  itself  in  motion  the  speed  of  the  point  in 
question  is  relative.  Linear  speed  is  expressed  in  linear  units,  per 
unit  of  time. 

22.  Angular  Speed  is  the  rate  of  turning  of  a  body  about  an  axis, 
or  the  rate  at  which  a  line  on  a  revolving  body  is  changing  direction, 
and  is  expressed  in  angular  units  per  unit  of  time. 

In  case  a  body  is  revolving  about  an  axis  outside  of  itself,  any 
point  in  the  body  has  only  linear  speed,  but  a  line,  real  or  imaginary, 
joining  the  point  to  the  axis  of  revolution  has  angular  speed,  also  a  line 
joining  any  two  points  on  the  body  has  angular  speed. 

23.  Uniform  and  Variable  Speed.     Speed  is  uniform  when  equal 
spaces  are  passed  over  in  equal  times,  however  small  the  intervals 
into  which  the  time  is  divided.     The  speed  in  this  case  is  the  space 
passed  over  in  a  unit  of  time,  and  if  s  represent  the  space  passed  over 
in  the  time  t,  the  speed  v  will  be 

'=f  (1) 

Speed  is  variable  when  unequal  spaces  are  passed  over  in  equal 
intervals  of  time,  increasing  spaces  giving  accelerated  motion  and 
decreasing  spaces  giving  retarded  motion.  The  speed,  when  variable, 
is  the  limit  of  the  space  passed  over  in  a  small  interval  of  time,  divided 
by  the  time,  when  these  intervals  of  time  become  infinitely  small.  If 
s  represents  the  space  passed  over  in  the  time  t,  then 

As 
v  =  limit  of  —  as  A£  approaches  zero, 

ds  fn^ 

•  -  ar  (2) 

The  uniform  speed  of  a  point  or  line  is  measured  by  the  number  of 
units  of  distance  passed  over  in  a  unit  of  time,  as  feet  per  minute, 
radians  per  second,  etc.  When  the  speed  is  variable  it  is  measured 
by  the  distance  which  would  be  passed  over  in  a  unit  of  time,  if  the 
point  or  line  retained  throughout  that  time  the  speed  which  it  had  at 
the  instant  considered. 

24.  Velocity  is  a  word  often  used  synonymously  with  speed,  al- 
though, accurately  speaking,   velocity  includes  direction  as  well  as 
speed.     The  linear  velocity  of  a  point  is  not  fully  defined  unless  the 
direction  in  which  it  is  moving  and  the  rate  at  which  it  is  moving  are 
both  known.     The  angular  velocity  of  a  line  would  be  defined  by 
giving  ite  angular  speed  and  the  direction  of  the  perpendicular  to  the 
plane  in  which  the  line  is  turning. 

25.  Linear  Acceleration  is  the  rate  of  change  of  linear  velocity. 
Since  velocity  involves  direction  as  well  as  rate  of  motion,  linear 


6  ELEMENTS  OF  MECHANISM 

acceleration  may  involve  a  change  in  speed  or  direction,  or  both. 
Any  change  in  the  speed  takes  place  in  a  direction  tangent  to  the 
path  of  the  point  and  is  called  tangential  acceleration,  while  a  change 
in  direction  takes  place  normal  to  the  path  and  is  called  normal 
acceleration.  This  must  never  be  confused  with  angular  acceleration 
which  will  be  discussed  later.  In  general  in  this  text  only  tangential 
acceleration  will  be  considered  and  it  will  be  understood  that  when 
the  word  acceleration  is  used  in  connection  with  linear  motion  it  is 
intended  to  refer  to  the  tangential  acceleration  unless  normal  accelera- 
tion is  definitely  mentioned.  The  following  example  will  make  clear 
the  meaning  of  terms  in  which  tangential  acceleration  is  expressed : 

If  a  body  is  moving  at  the  rate  of  1  ft.  per  second  at  the  end  of  the 
first  second,  3  ft.  per  second  at  the  end  of  the  second  second,  5  ft. 
per  second  at  the  end  of  the  third  second,  etc.,  the  speed  is  increasing 
at  the  rate  of  2  ft.  per  second  each  second.  Its  acceleration  is  two  feet 
per  second  each  second. 

Acceleration  may  be  either  positive  or  negative.  If  the  speed  is 
increasing  the  acceleration  is  positive;  if  the  speed  is  decreasing  the 
acceleration  is  negative.  If  the  speed  changes  the  same  amount 
each  second  the  acceleration  is  uniform,  but  if  the  speed  changes  by 
different  amounts  at  different  times  the  acceleration  is  variable.  If 
Av  represents  the  change  in  speed  in  the  time  A£,  then,  if  the  accelera- 
tion a  is  uniform 

_«-£  (3) 

When  the  acceleration  is  variable 

a  =  limit  of  —  as  At  approaches  zero 

dv  ,4^ 

or  a  =  dt'  {' 

26.  Angular  Acceleration  is  rate  of  change  of  angular  velocity.  In 
this  case,  as  in  the  case  of  linear  acceleration,  a  change  in  either  speed 
or  direction  of  rotation,  or  both,  may  be  involved.  For  example,  if 
a  line  is  turning  in  a  plane  with  a  varying  angular  speed  it  has  angu- 
lar acceleration  which  may  be  positive  or  negative;  or  if  the  direction 
of  the  plane  of  rotation  is  changing  the  line  also  has  angular  accelera- 
tion. Unless  otherwise  stated  angular  acceleration  in  this  text  will 
be  understood  to  refer  to  change  in  angular  speed.  Angular  accelera- 
tion is  expressed  in  angular  units  change  in  speed  per  unit  time  (such 
as  radians,  degrees,  or  revolutions  per  minute  each  minute).  Equa- 
tions (3)  and  (4)  will  apply  to  angular  acceleration  if  a  and  v  are 
expressed  in  angular  units. 


Variable  motion 


INTRODUCTION  7 

27.  Kinds  of  Motion.  From  the  preceding  discussion  it  is  evident 
that  motion,  whether  absolute  or  relative,  may  be  classified  as  follows : 

Uniform  motion  —  Acceleration  zero. 

Acceleration  constant 
Acceleration  variable 
Acceleration  constant  part  of  the 
time  and  variable  part  of  the  time. 

A  body  having  uniform  motion  travels  without  change  of  speed, 
if  any  normal  acceleration  is  neglected.  For  example,  if  a  block  is 
sliding  on  guides  a  distance  of  one  inch  each  second,  or  a  wheel  turn- 
ing constantly  at  the  rate  of  100  revolutions  per  minute,  each  would 
be  said  to  have  uniform  motion. 

The  most  familiar  example  of  motion  having  constant  acceleration 
is  that  of  a  body  falling  freely  under  the  action  of  gravity.  A  weight 
dropped  from  a  height  will  have  at  the  start  a  zero  speed.  Under 
the  action  of  the  constant  force  of  gravity  its  speed  will  increase  at  a 
constant  rate  such  that  at  the  end  of  one  second  the  body  will  be 
moving  at  the  rate  of  about  32.2  ft.  per  second  and  it  will  have 
dropped  16.1  ft.  At  the  end  of  two  seconds  it  will  have  a  speed  of 
64.4  ft.  per  second  and  will  have  dropped  48.3  ft.  If  the  constant 
force  acting  were  of  different  magnitude  the  amount  of  acceleration 
would  be  different  but  it  would  still  be  constant.  The  space  moved 
over  in  successive  intervals  of  time  by  a  body  having  constant  accelera- 
tion will  be  in  the  ratio  of  the  odd  numbers,  1,  3,  5,  7,  9,  etc. 

In  the  case  of  machine  parts  having  variable  acceleration  the 
variation  does  not  usually  follow  any  simple  law. 
One  form  of  motion  with  variable  acceleration, 
however,  occurs  not  infrequently,  and  is  known  as 
simple  harmonic  motion.  If  a  point,  as  R,  Fig.  1, 
moves  with  uniform  speed  around  the  circumference  FIG  l 

of  a  semicircle  and  another  point  T  moves  across 
the  diameter  in  the  same  length  of  time,  the  speed  of  T  varying  so 
that  it  will  always  be  at  the  foot  of  a  perpendicular  let  fall  from  R  to 
the  line  A  B.  then  T  is  said  to  have  harmonic  motion. 

The  relation  between  the  displacement  AT  of  the  point  T  from  its 
initial  position  and  the  angle  0  is  expressed  by  the  equation 

AT  =  OA  -  OR  cos  6  =  OR  (1  -  cos  0).  (5) 

The  acceleration  of  the  point  T  may  be  deduced  as  follows: 

Let  s  =  AT;  co  =  angular  speed  of  OR  in  radians  per  second;  t  =  time 
required  for  line  OR  to  move  through  angle  6,  therefore  6  =  ut'  v  =  speed 


8  ELEMENTS  OF  MECHANISM 

of  T  in  linear  units  per  second,  a  =  acceleration  in  similar  units  per 
second  each  second. 
Then  from  equation  (5) 

»  =  OR  -  OR  cos  6  =  OR  -  OR  cos  cot. 

From  equation  (2)      v  =  -57  =  &OR  sin  at, 

dv      d?s 
a  =  di  =  dt* 

OT 
but  cos  <*>t  =  ^  • 

OT 
Therefore  a  =  MR  X  ^%  =  co2OT.  (5a) 

C//L 

Therefore  the  acceleration  is  proportional  to  the  distance  of  the  mov- 
ing point  from  the  center  of  its  path.  When  T  is  approaching  0  its 
acceleration  is  positive  and  when  receding  from  0  the  acceleration  is 
negative. 

The  arrangement  of  parts  in  a  mechanism  may  be  such  that  one  or 
more  of  the  pieces  has  an  increasing  speed  at  the  beginning,  then  moves 
at  uniform  speed  during  the  greater  part  of  the  motion,  and  has  de- 
creasing speed  at  the  end. 

28.  Modification  of  Motion.  In  the  action  of  a  mechanism  the 
motion  of  the  follower  may  differ  from  that  of  the  driver  in  kind,  in 
speed,  in  direction,  or  in  all  three.  As  the  paths  of  motion  of  the 
driver  and  follower  depend  upon  the  connections  with  the  frame  of  the 
machine,  the  change  of  motion  in  kind  is  fixed,  and  it  only  remains  to 
determine  the  relations  of  direction  and  speed  throughout  the  motion. 
The  laws  governing  the  changes  in  direction,  and  speed  can  be  deter- 
mined by  comparing  the  movements  of  the  two  pieces  at  each  instant 
of  their  action,  and  the  mode  of  action  will  fix  the  laws.  Therefore, 
whatever  the  nature  of  the  combination,  if  it  is  possible  to  determine, 

throughout  the  motion  of  the 
driver  and  follower,  the  speed 
ratio  and  directional  relation, 
the  analysis  will  be  complete. 

Either  the  speed  ratio  or  the 
directional  relation  may  vary, 
or  remain  the  same  throughout 
the  action  of  the  two  pieces. 

•pIG  2  29.   Pairs  of  Elements.     In 

order  that  a  moving  body,  as  A 

(Fig.  2),  may  remain  continually  in  contact  with  another  body  B,  and 
at  the  same  time  move  in  a  definite  path,  B  would  have  a  shape  which 


INTRODUCTION 


9 


could  be  found  by  allowing  A  to  occupy  a  series  of  consecutive  positions 
relative  to  B,  and  drawing  the  envelope  of  all  these  positions.  Thus, 
if  A  were  of  the  form  shown  in  the  figure,  the  form  of  B  would  be  that 
of  a  curved  channel.  Therefore,  in  order  to  compel  a  body  to  move  in 
a  definite  path,  it  must  be  paired  with  another,  the  shape  of  which  is 
determined  by  the  nature  of  the  relative  motion  of  the  two  bodies. 

30.  Closed  or  Lower  Pair.  If  one  element  not  only  forms  the 
envelope  of  the  other,  but  encloses  it,  the  forms  of  the  elements  being 
geometrically  identical,  the  one  being  solid  or  full,  and  the  other 
being  hollow  or  open,  we  have  what  may  be  called  a  closed  pair,  also 
called  a  lower  pair.  In  such  a  pair,  surface  contact  exists  between  the 
two  members. 

On  the  surfaces  of  two  bodies  forming  a  closed  pair,  coincident  lines 
may  be  supposed  to  be  drawn,  one  on  each  surface;  and  if  these  lines 
are  of  such  form  as  to  allow  them  to  move  along  each  other,  that  is,  allow 
a  certain  motion  of  the  two  bodies  paired,  three  forms  only  can  exist : 

1°  A  straight  line,  which  allows  straight  translation  (Fig.  3). 

2°  Among  plane  curves,  or  curves  of  two  dimensions,  a  circle, 
which  allows  rotation,  or  revolution  (Figs.  4  and  5). 

3°  Among  curves  of  three  dimensions,  the  helix,  which  allows  a 
combination  of  rotation  and  straight  translation  (Fig.  6). 


FIG.  3 


FIG.  4 


FIG.  6 

31.  Higher  Pairs.     The  pair  represented  in  Fig.  2  is  not  closed,  as 
the  elementary  bodies  A  and  B  do  not  enclose  each  other  in  the  above 
sense.     Such  a  pair  is  called  a  higher  pair  and  the  contact  between  the 
elements  is  along  lines  only. 

32.  Incomplete  Pairs  of  Elements.     Hitherto  it  has  been  assumed 
that  the  reciprocal  restraint  of  two  elements  forming  a  pair  was  com- 
plete, i.e.,  that  each  of  the  two  bodies,  by  the  rigidity  of  its  material 


10  ELEMENTS  OF  MECHANISM 

and  the  form  given  to  it,  restrained  the  other.  In  certain  cases  it  is 
only  necessary  to  prevent  forces  having  a  certain  definite  direction 
from  affecting  the  pair,  and  in  such  cases  it  is  no  longer  absolutely 
necessary  to  make  the  pair  complete;  one  element  can  then  be  cut 
away  where  it  is  not  needed  to  resist  the  forces. 

The  bearings  for  railway  axles,  the  steps  for  water-wheel  shafts,  the 
ways  of  a  planer,  railway  wheels  kept  in  contact  with  the  rails  by  the 
force  of  gravity  are  all  examples  of  incomplete  pairs  in  which  the  ele- 
ments are  kept  in  contact  by  external  forces. 

33.  Inversion  of  Pairs.  In  Fig.  3  if  B  is  the  fixed  piece  all  points 
on  A  move  in  straight  lines.  If  A  were  the  fixed  piece  all  points  in  B 
would  move  in  straight  lines.  That  is,  the  absolute  motion  of  the 
moving  piece  is  the  same,  whichever  piece  is  fixed.  The  same  state- 
ment holds  true  of  the  pairs  shown  in  Figs.  4,  5  and  6. 

This  exchange  of  the  fixedness  of  an  element  with  its  partner  is 
called  the  inversion  of  the  pair,  and  in  the  case  of  any  closed  or  lower 
pair  it  does  not  effect  either  the  absolute  or  the  relative  motion. 

In  the  pairs  shown  in  Figs.  2   and  7,  both  of  which  are  higher  pairs, 
the  relative  motion  of  A  and  B  is  the  same  when  A  is  fixed  as  when  B  is 
fixed.     The  absolute  motion  of  A  when  B  is  fixed  is 
not  the  same  as  the  absolute  motion  of  B  when  A  is 
A      ]  fixed. 

^3       34.   Bearings.     The  word  bearing  is  applied,  in 
PJ     ^  general,  to  the  surfaces  of  contact  between  two 

pieces  which  have  relative  motion,  one  of  which 
supports  or  partially  supports  the  other.  One  of  the  pieces  may  be 
stationary,  in  which  case  the  bearing  may  be  called  a  stationary 
bearing;  or  both*  pieces  may  be  moving. 

The  bearings  may  be  arranged,  according  to  the  relative  motions 
they  will  allow,  in  three  classes: 

1°  For  straight  translation  the  bearings  must  have  plane  or  cylin- 
drical surfaces,  cylindrical  being  understood  in  its  most  general 
sense.  If  one  piece  is  fixed  the  surfaces  of  the  moving  pieces  are  called 
slides ;  those  of  the  fixed  pieces,  slides  or  guides. 

2°  For  rotation,  or  turning,  the  bearings  must  have  surfaces  of 
revolution,  as  circular  cylinders,  cones,  conoids  or  flat  disks.  The 
surface  of  the  solid  or  full  piece  is  called  a  journal,  neck,  spindle  or 
pivot;  that  of  the  hollow  or  open  piece,  a  bearing,  gudgeon,  pedestal, 
plumber-block,  pillow-block,  bush  or  step. 

3°  For  translation  and  rotation  combined,  or  helical  motion,  they 
must  have  a  helical  or  screw  shape,  Here  the  full  piece  is  called  a 
screw  and  the  open  piece  a  nut, 


INTRODUCTION 


11 


35.  Collars  and  Keys.  It  is  very  often  the  case  that  pulleys  or 
wheels  are  to  turn  freely  on  their  cylindrical  shafts  and  at  the  same 
time  have  no  motion  along  them.  For  this  purpose,  rings  or  collars 
(Fig.  8a)  are  used;  the  collars  D  and  E,  held  by  set  screws,  prevent 
the  motion  of  the  pulley  along  the  shaft  but  allow  it  free  rotation. 
Sometimes  pulleys  or  couplings  must  be  free  to  slide  along  their 


FIG.  8 


shafts,  but  at  the  same  time  must  turn  with  them;  they  must  then 
be  changed  to  a  sliding  pair.  This  is  often  done  by  fitting  to  the 
shaft  and  pulley  or  sliding  piece  a  key  C  (Fig.  8b),  parallel  to  the 
axis  of  the  shaft.  The  key  may  be  made  fast  to  either  piece,  the 
other  having  a  groove  in  which  it  can  slide  freely.  The  above  ar- 
rangement is  very  common,  and  is  called  a  feather  and  groove  or 
spline,  or  a  key  and  key  way. 


CHAPTER  II 
REVOLVING  AND   OSCILLATING  BODIES 

36.  Revolving  Bodies.*     One  of  the  most  common  motions  in  ma- 
chinery is  revolution  or  rotation  about  an  axis.     The  rotating  body 
may  be  a  cylinder,  or  cone,  or  a  piece  of  irregular  form.     The  shaft  or 
element  upon  which  the  rotating  piece  is  supported  may  turn  with  it, 
being  itself  supported  in  bearings  and  restrained  from  moving  endwise 
by  collars;   or  the  shaft  may  be  held  stationary  and  the  piece  turn  on 
the  shaft.     An  example  of  the  first  case  is  shown  in  Fig.  9  and  of  the 
second  case  in  Fig.  10. 

37.  Angular  Speed.     Suppose  some  force  is  applied  to  the  shaft  in 
Fig.  9  so  that  it  is  caused  to  turn  around,  say,  75  times  in  a  minute. 


Collar 


FIG.  9  FIG.  10 

If  the  wheel  is  fast  to  the  shaft,  the  wheel  will  turn  around  75  times 
in  a  minute.  It  would  be  said,  then,  that  the  wheel  makes  75  revolu- 
tions per  minute  (usually  abbreviated  thus:  75  r.p.m.)  It  is  common 
practice  to  speak  of  the  angular  speed  or  speed  of  turning  in  terms  of 
revolutions  per  unit  of  time,  usually  per  minute  or  second. 

Another  unit  sometimes  used  for  measuring  angular  motion  is  the 
angle  called  the  radian.  This  is  the  angle  which  is  subtended  by  the 
arc  of  a  circle  equal  in  length  to  the  radius.  Since  the  radius  is  con- 
tained in  the  circumference  of  a  circle  2  TT  times,  there  must  be  2  TT 
radians  in  360°,  or  one  radian  is  equal  to  360°  ^  2  TT  =  57°  17'  42". 

If  a  revolving  wheel  turns  once  per  minute,  its  angular  speed  as  we 
have  already  seen  is  1  r.p.m.,  and  since  one  revolution  of  the  wheel  causes 

*  Throughout  this  book  the  words  revolve  and  turn  will  refer  to  turning  about  any 
axis,  whether  within  or  outside  the  body  in  question,  while  rotate  will  refer  only  to 
turning  about  an  axis  passing  through  the  body. 

12 


REVOLVING  AND  OSCILLATING  BODIES  13 

any  radial  line  on  the  wheel  to  sweep  over  360°  or  2  TT  radians,  the  angu- 
lar speed  of  the  wheel  is  2  TT  radians  per  minute.  Now,  if  the  wheel 
turns  N  times  per  minute,  the  angular  speed  is  N  r.p.m.  or  2  nN  radians 
per  minute.  That  is, 

Angular  speed  in  radians  =  2ir  X  number  of  revolutions.        (6) 

Example  1.     If  a  wheel  turns  90  r.p.m.  its  angular  speed  in  radians  is 
2  TT  90  =  565.5  radians  per  minute. 

Referring  to  Fig.  11,  let  the  body  M  be  rigidly  attached  to  an  arm 
which  is  turning  around  the  axis  C,  the  arm  and  M  revolving  together. 
Then  the  lines  CA  and  CB  which  join  any  two  points  A  and  B  to  the 
axis  have  angular  speed  about  C  and  since  the  entire  body  is  rigid  and 
the  angle  ACB  is  constant,  CA  and  CB  each  have  the  same  angular 
speed  as  the  arm.  Moreover,  since,  as  the  body  revolves,  the  line  AB 
constantly  changes  direction,  it  may  also  be  said  to  have  angular  speed, 
which,  in  this  case,  is  the  same  as  that  of  the  lines  CA  and  CB. 

If  M  is  not  rigidly  attached  to  the  arm  but  is  rotating  on  the  axis  S 
which  is  carried  by  the  arm,  as  in  Fig.  12,  the  lines  CA,  CB  and  AB  will 


FIG.  11  FIG.  12 

no  longer  necessarily  have  the  same  angular  speed  since  the  angles 
turned  through  in  a  given  time  by  these  lines  depend  not  only  on  the 
speed  at  which  the  arm  is  turning  about  C  but  also  upon  the  speed  at 
which  M  is  turning  about  the  axis  S. 

38.  Linear  Speed  of  a  Point  on  a  Revolving  Body.  Consider  a 
point  A  on  the  circumference  of  the  wheel  in  Fig.  9.  While  the  wheel 
turns  once,  A  travels  over  the  circumference  of  a  circle  of  the  same 
diameter  as  the  wheel,  or  it  travels  2  irR  in.  if  the  radius  of  the  wheel 
is  R  in.  Then,  if  the  wheel  turns  N  times  in  a  unit  of  time,  A  travels 
over  the  circle  N  times  in  a  unit  of  time.  Therefore,  the  linear  speed 
of  the  point  A  is  2  irRN  in.  per  unit  of  time.  Writing  this  in  the  form 
of  an  equation, 

Linear  speed  of  A  =  2  irRN.  (7) 


14  ELEMENTS  OF  MECHANISM 

Example  2.  Suppose  the  wheel  is  12  in.  in  diameter  and  turns  40  times  per 
minute.  The  speed  of  A  would  be 

2  TT  6  X  40  in.  per  minute  =  1506  in.  per  minute. 

From  equation  (6)  it  appears  that  the  angular  speed  is  equal  to  2  irN 
radians;  so  that,  dividing  equation  (7)  by  equation  (6) 

Linear  speed  of  A 2  irRN  _  R 

Angular  speed  of  wheel  in  radians       2  -n-N       1 
or, 

Linear  speed  of  a  point  on  a  revolving  body  =  angular  speed  of  body 

in  radians  X  distance  of  the  point  from  the  center.  (8) 

39.  Speed  Ratio  of  Points  at  Different  Distances  from  Axis.  If 
another  point  B  is  chosen  on  the  side  of  the  wheel  at  a  distance  Ri 
from  the  center,  it  can  be  shown  in  the  same  way  that  Eq.  (7)  was 

derived,  that 

Linear  speed  of  B  =  2irRiN .  (9) 

Now,  dividing  Eq.  (7)  by  Eq.  (9), 

Linear  speed  of  A  _  2  irRN 
Linear  speed  of  B   2-n-RiN 

m,   „  Linear  speed  of  A   R  ,  . 

Therefore,  K     ,    .  p  =  ^  '  (10) 

Linear  speed  of  B      Ri 

Thus,  the  linear  speeds  of  two  points  on  a  revolving  wheel  are  directly 
proportional  to  the  distances  of  the  points  from  the  center  about  which  the 
wheel  is  turning. 

The  linear  speed  of  a  point  on  the  circumference  of  a  wheel  is  often 
spoken  of  as  the  periphery-speed  or  surface  speed  of  the  wheel. 

Take  another  case,  that  of  two  wheels 
fast  to  the  same  shaft  as  shown  in  Fig.  13. 
The  weight  P  is  supposed  to  be  hung  from  a 
steel  band  which  is  wound  on  the  outside 
of  wheel  A  and  the  weight  W  from  another 
steel  band  wound  on  the  outside  of  wheel  B. 
Suppose  that  the  shaft  starts  to  turn  in  the 
direction  shown  by  the  arrow.  Then  the 
.p  13  band  which  supports  P  will  be  paid  out, 

that  is,  will  unwind,  at  a  speed  equal  to  the 

periphery  speed  of  A  and  the  weight  P  will  descend  at  that  speed. 
At  the  same  time  the  other  band  will  be  winding  on  to  the  wheel  B 
and  the  weight  W  will  be  rising  at  a  speed  equal  to  the  periphery  speed 
of  B.  If  N  represents  the  number  of  turns  per  second  of  the  shaft, 


REVOLVING  AND  OSCILLATING  BODIES  15 

R  the  radius  of  A,  RI  the  radius  of  B,  then  the  speed  of  P  =  2irRN 
and  speed  of  W  =  2  nRiN,  or 

Speed   P      R 


Speed 


(ID 


which  is  the  same  equation  found  when  both  points  were  on  the  same 
wheel. 

Example  3.  Let  the  diameter  of  A  —  12  in.  and  the  diameter  of  B  =  8  in.  and 
let  the  shaft  turn  1|  times  per  second.  Then 

Speed  P  =  27rX6Xl£  =  56.55  in.  per  second. 
Speed  T7=2?rX4XH  =  37.70  in.  per  second. 

Now,  according  to  Eq.  11,  these  speeds  should  be  in  the  ratio  of  6  to  4  or  1.5  to  1, 
and  if  56.55  is  divided  by  37.7,  the  result  is  equal  to  1.5  except  for  the  slight  error 
due  to  carrying  the  figures  to  the  nearest  hundredth  of  an  inch. 

40.  Relation  between  Forces  and  Speeds.  Suppose  that  in  Fig.  13 
it  is  assumed  that  there  is  no  friction  and  that  the  weights  of  P  and 
W  are  such  that,  if  the  shaft  is  at  rest,  the  weights  will  just  balance 
each  other,  or,  if  the  shaft  is  caused  to  start  turning  in  a  given  direction, 
the  weights  will  allow  it  to  keep  turning  at  a  uniform  speed. 

The  work  done  by  a  force  is  equal  to  the  force,  expressed  in  units 
of  force,  multiplied  by  the  distance  through  which  the  force  acts,  ex- 
pressed in  linear  units,  provided  the  motion  takes  place  in  the  direc- 
tion in  which  the  force  acts.  Now,  if  friction  is  neglected,  the  work 
obtained  from  a  machine  must  equal  the  work  put  into  it.  Hence,  in 
Fig.  13,  if  the  falling  weight  P  be  considered  as  the  force  driving  the 
machine,  the  work  put  into  the  machine  is  the  weight  P  multiplied  by 
the  distance  P  falls.  The  work  obtained  from  the  machine  is  the 
weight  of  W  multiplied  by  the  distance  W  is  raised. 

Then  the  weight  of  P  (in  pounds  or  other  weight  units)  multiplied 
by  the  distance  P  moves  in  a  given  length  of  time  is  equal  to  the  weight 
of  W  multiplied  by  the  distance  it  moves  in  the  same  time,  the  units 
of  weight  and  distance  being  the  same  in  both  cases.  If  the  shaft  is 
assumed  to  make  N  turns,  the  distances  moved  by  P  and  W  are  2  irRN 
and  2  irRiN,  respectively. 


Therefore,  P  X  2  irRN  =  W  X  2  nR^,  (12) 

Weight  of  P_fli. 
Weight  of  W  ~  R 

Since  from  Eq.  (11)  the 

Linear  speed  of  P  _  R 
Linear  speed  of  W      RI 


16 


ELEMENTS  OF  MECHANISM 


Therefore, 


(14) 


Weight  of  P  _  Speed  of  Wm 

Weight  of  W  ~  Speed  of  P  ' 
41.   Cranks  and  Levers.     A  crank  may  be  defined  in  a  general 
way  as  an  arm  rotating  or  oscillating  about  an  axis.     It  may  be 

thought  of  as  a  piece  cut  out  of 
a  wheel  or  disk,  as  suggested  by 
the  dotted  circle  in  Fig.  14,  and 
the  laws  for  revolving  wheels 
apply  equally  to  cranks.  When 
two  cranks  are  rigidly  connected 
to  each  other  the  name  lever  is 
often  applied  to  the  combination, 
particularly  when  the  motion  is 
oscillating  over  a  relatively  small 


\ 


FIG.  14 


angle. 


In  Fig.  15  the  two  arms  of  the  lever  are  shown  at  an  angle  of  180° 
with  each  other.  This  condition  does  not  necessarily  hold,  however, 
for  the  two  arms  may  make  any  angle  with  each  other  from  180°  as  in 
Fig.  15  down  to  0°  as  in  Fig.  16.  When  the  angle  between  the  two 
arms  is  less  than  90°  as  in  Figs.  16  and  17  it  is  often  called  a  bell  crank 


FIG.  16 


FIG.  17 

lever,  and  when  the  angle  is  more  than  90°  as  in  Figs.  15  and  18  it  is 
often  called  a  rocker.  These  terms,  however,  are  used  rather  loosely 
and  somewhat  interchangeably. 

In  all  these  cases  the  following  equation  holds  true: 
Linear  speed  A  _  Distance  of  A  from  axis 
Linear  speed  B  ~  Distance  of  B  from  axis 
*  It  must  be  always  borne  in  mind  that  any  equation  such  as  Eq.  (14)  does 
not  take  friction  into  account. 


(15) 


REVOLVING  AND  OSCILLATING  BODIES 


17 


The  two  lever  arms  may  be  in  the  same  plane  as  in  Figs.  15  to  18  or 
they  may  be  attached  to  the  same  shaft  but  lie  in  different  planes  as 
in  Fig.  19. 

42.  Motion  from  Levers.  It  is  often  necessary  to  transfer  some 
small  motion  from  one  line  to  another.  Three  cases  will  be  considered 
which  depend  on  the  relative  positions  of  the  lines  of  motion: 

1°  Parallel  lines. 

2°  Intersecting  lines. 

3°  Lines  neither  parallel  nor  intersecting. 

The  first  case  is  an  application  of  the  form  of  lever  shown  in  Figs.  15 
and  16,  and  also  Fig.  19  if  the  arms  CA  and  CB  are  in  the  same  plane 

C 


FIG.  18 

passing  through  the  axis  of  the  shaft.     The  motions  of  A  and  B  are 
directly  proportional  to  their  distances  from  the  axis  C.     In  Fig.  15, 


FIG.  20a 


A  and  B  are  always  moving  in  opposite  directions,  while,  in  Fig.  16, 
A  and  B  move  in  the  same  direction. 

In  the  second  case  a  bent  lever  is  used  of  the  type  shown  in  Figs. 
17  and  18.     Referring  to  Fig.  20a,  let  it  be  assumed  that  a  lever  is  to 


18 


ELEMENTS  OF  MECHANISM 


be  laid  out  which  will  give  motion  along  AD  bearing  a  known  ratio  to 
that  along  BD. 

Draw  the  line  DC,  dividing  the  angle  ADB  into  two  angles  ADE 
and  BDE  whose  sines  are  directly  proportional  to  the  motions  required 
along  AD  and  BD  respectively. 

This  may  be  done  by  erecting  perpendiculars  MN  and  PT  on  AD 
and  BD  in  the  ratio  of  the  required  motions  along  those  lines,  and  draw- 
ing through  their  extremities  N  and  T  lines  parallel  to  AD  and  BD 
respectively;  the  intersection  of  these  lines  at  E  determines  the  line 


FIG.  20b 

DE.  Choose  any  point  C  in  DE,  and  drop  the  perpendiculars  CA 
and  CB  on  AD  and  BD  respectively;  then  ACB  is  the  bell-crank  lever 
required.  As  the  lever  moves 

Linear  speed  of  A  _  AC  _  sin  CD  A 
Linear  speed  of  B  ~  BC  ~  sin  CDB 

It  is  evident  that,  for  a  small  angular  motion,  the  movements  in 
AD  and  BD  are  very  nearly  rectilinear,  and  they  will  become  more  and 
more  so  the  farther  C  is  removed  from  the  point  D. 

Any  slight  motion  that  may  occur  perpendicular  to  the  lines  AD 
and  BD  may  be  provided  for  by  the  connectors  used.  It  is  to  be 
noticed,  however,  that  for  a  given  motion  on  the  lines  AD  and  BD 
these  perpendicular  movements,  or  deviations,  are  less  when  the  lever- 
arms  vibrate  equal  angles  each  side  of  the  positions  which  they  occupy 
when  perpendicular -to  the  lines  of  motion,  and  they  should  always  be 


REVOLVING  AND  OSCILLATING  BODIES 


19 


arranged  to  so  vibrate.  By  moving  the  point  C  nearer  to  D,  at  the 
same  time  keeping  the  lever-arms  the  same,  this  perpendicular  devia- 
tion could  be  disposed  equally  on  each  side  of  the  lines  of  motion 
which  is  advisable  especially  in  cases  where  the  deviation  is  allowed 


FIG.  21 

for  by  the  spring  of  the  connecting  piece.  This  is  shown  in  position 
AiCiBi,  Fig.  20b.  In  Fig.  20  it  will  be  seen  that  A  and  B  move  in 
opposite  directions,  while  Fig.  21  shows  the  result  if  A  and  B  are  to 
move  in  the  same  direction. 


FIG.  22 

In  the  third  case  a  lever  of  the  type  of  Fig.  19  would  be  used  with 
the  arms  CA  and  CB  making  the  proper  angle  with  each  other.    Let 


20 


ELEMENTS  OF  MECHANISM 


BD  be  the  line  along  which  B  is  to  give  motion  and  AD  the  line  along 
which  A  is  to  give  motion.  Let  BD  lie  in  the  plane  XY  and  AD  lie  in 
the  plane  VW.  To  find  the  position  of  the  line  DC  which  is  the  trace 
of  the  plane  containing  the  axis  of  the  shaft,  assume  the  plane  VW  to 
be  moved  to  the  left  until  it  coincides  with  XY.  Then  lay  out  the 

lever  in  the  left  elevation  as  described 
for  Fig.  20.  Next  assume  the  plane 
VW  moved  back  to  its  proper  posi- 
tion, carrying  the  arm  CA  with  it. 

43.  Effective    Lever    Arms.      In 
the    case    of   a   lever   in  a  position 


M 


FIG.  23 


FIG.  24 


such  as  indicated  in  Figs.  23  or  24,  the  effect  is  the  same,  for  the 
instant,  as  if  the  weights  P  and  PI  were  attached  to  the  lever  MCN, 
whose  arms  are  found  by  drawing  perpendiculars  from  the  axis  C  to 
the  line  of  action  of  the  forces  exerted  by  the  weights  P  and  PI. 
The  perpendiculars  CM  and  CN  may  be  called  the  effective  lever 
arms  or  moment  arms  of  the  weights. 


CHAPTER   III 
BELTS,   ROPES   AND   CHAINS 

44.  Flexible  Connectors.  If  the  wheel  A,  Fig.  25,  is  turning  at  a 
certain  angular  speed  about  the  axis  S  its  outer  surface  will  have  a  linear 
speed  dependent  upon  the  angular  speed  and  the  diameter  of  A.  (See 
§38.) 

If  a  flexible  band  is  stretched  over  A}  connecting  it  with  another 
wheel  B  and  there  is  sufficient  friction  between  the  band  and  the  sur- 
faces of  the  wheels  to  prevent 
appreciable  slipping,  then  the 
band  will  move  with  a  linear 
speed  approximately  equal  to 
the  surface  speed  of  A,  and  will 
impart  approximately  the  same 
linear  speed  to  the  surface  of  B, 
thus  causing  B  to  turn.  The 
wheels  may  be  on  axes  which  pIG 

are  parallel,  intersecting,  or 

neither  parallel  nor  intersecting.     Flexible  connectors  may  be  divided 
into  three  general  classes: 

1°  Belts  made  of  leather,  rubber,  or  woven  fabrics  are  flat  and 
thin,  and  require  pulleys  nearly  cylindrical  with  smooth  surfaces. 
Flat  ropes  may  be  classed  as  belts. 

2°  Cords  made  of  catgut,  leather,  hemp,  cotton  or  wire  are  nearly 
circular  in  section  and  require  either  grooved  pulleys  or  drums  with 
flanges.  Rope  gearing,  either  by  cotton  or  wire  ropes,  may  be  placed 
under  this  head. 

3°  Chains  are  composed  of  links  or  bars,  usually  metallic,  jointed 
together,  and  require  wheels  or  drums  either  grooved,  notched,  or 
toothed,  so  as  to  fit  the  links  of  the  chain. 

For  convenience  the  word  band  may  be  used  as  a  general  term  to 
denote  all  kinds  of  flexible  connectors. 

Bands  for  communicating  continuous  motion  are  endless. 

Bands  for  communicating  reciprocating  motion  are  usually  made 
fast  at  their  ends  to  the  pulleys  or  drums  which  they  connect. 

21 


22  ELEMENTS  OF  MECHANISM 

45.  Pitch  Surface  and  Line  of  Connection.     Fig.  26  represents  the 
edge  view  of  a  piece  of  a  belt  before  being  wrapped  around  the  pulley. 
If  it  is  assumed  that  there  are  no  irregularities  in  the  make  up  of  the 
belt  the  upper  surface  o  is  parallel  to  and  equal  in  length  to  the  sur- 
face i.     When  this  same  belt  is  stretched  around  a  pulley,  as  in  Fig. 
27,  the  surface  i  is  drawn  firmly  against  the  surface  of  the  pulley  while 
the  surface  o  bends  over  a  circle  whose  radius  is  greater  than  that  of 
the  surface  of  the  pulley  by  an  amount  equal  to  the  belt  thickness  2  p. 

The  outer  part  of  the  belt  must 
therefore  stretch  somewhat  and 
the  inner  part  compress.  There 
will  be  some  section  between  i 
and  o  which  is  neither  stretched 
nor  compressed  and  the  name 
neutral  section  may  be  given  to 
this  part  of  the  belt.  In  the  case  of  a  flat  belt  the  neutral  section 
may  be  assumed  to  be  half  way  between  the  outer  and  inner  surfaces. 
An  imaginary  cylindrical  surface  around  the  pulley,  to  which  the 
neutral  section  of  the  belt  is  tangent,  is  the  pitch  surface  of  the 
pulley,  the  radius  of  this  being  the  effective  radius  of  the  pulley. 
A  line  in  the  neutral  section  of  the  belt  at  the  center  of  its  width  is 
the  line  of  connection  between  two  pulleys  and  is  tangent  to  the  pitch 
surfaces,  and  coincides  with  a  line  in  each  pitch  surface  known  as  the 
pitch  line. 

46.  Speed  Ratio  and  Directional  Relation  of  Shafts  Connected  by  a 
Belt.     In  Fig.  25  let  the  diameter  of  the  pulley  A  be  D  inches,  the  di- 
ameter of  B  be  Di  inches  and  the  half  thickness  of  belt  =  p.    Also  let 
N  represent  the  r.p.m.  of  S,  and  JVi  =  r.p.m.  of  Si. 

Then,  from  equation  (7), 

Linear  speed  of  pitch  surface  of  A  =  irN  (D  +  2  p), 
and  Linear  speed  of  pitch  surface  of  B  =  irNi  (Di  +  2  p) . 

If  the  belt  speed  is  supposed  to  be  equal  to  the  speed  of  the  pitch 
surfaces  of  the  pulleys 

irN  (D  +  2P)  =*Ni(Dl  +  2P), 

N       Dl  +  2P 
Nl  =  D  +  2-p' 

That  is,  the  angular  speeds  of  the  shafts  are  in  the  inverse  ratio  of  the 
effective  diameters  of  the  pulleys,  and  this  ratio  is  constant  for  circular 
pulleys. 


BELTS,  ROPES  AND  CHAINS  23 

As  the  thickness  of  belts  generally  is  small  as  compared  with  the 
diameters  of  the  pulleys,  it  may  be  neglected. 
The  speed  ratio  will  then  become 

N       Dl 


which  is  the  equation  almost  always  used  in  practical  calculations. 

Example  4.  Assume  that  a  shaft  A  makes  360  r.p.m.  On  A  is  a  pulley  24  ins. 
in  diameter  belted  to  a  pulley  36  ins.  in  diameter  on  another  shaft  B.  To  find  speed 
of  shaft  B. 

From  Eq.  (17) 

Speed  of  A  _  Diam.  of  pulley  on  B 
Speed  of  B      Diam.  of  pulley  on  A 

Substituting  the  known  values,  this  equation  becomes 

360        =36 
Speed  of  B      24* 

Therefore,  Speed  of  B  =  f|  X  360  =  240  r.p.m. 

Example  5.  Suppose  a  shaft  A  making  210  r.p.m.  is  driven  by  a  belt  from  a  30- 
in.  pulley  on  another  shaft  B  which  makes  140  r.p.m.',  to  find  the  size  of  the  pulley 
on  A. 

Using  the  principle  of  Eq.  (17) 

Speed  of  A  _  Diam.  of  pulley  on  B 
Speed  of  B      Diam.  of  pulley  on  A 

Therefore,  ™  .  f  or  ,  -  "X™  -  20  in, 

Then  a  20-in.  pulley  is  required  on  A. 

The  relative  directions  in  which  the  pulleys  turn  depend  upon  the 
manner  in  which  the  belt  is  put  on  the  pulleys.  The  belt  shown  in  Fig. 
25  is  known  as  an  open  belt  and 
the  pulleys  turn  in  the  same 
direction,  as  suggested  by  the 
arrows.  The  belt  shown  in  Fig. 
28  is  known  as  a  crossed  belt 
and  the  pulleys  turn  in  opposite 
directions  as  indicated. 

47.   Kinds  of  Belts.    The  ma- 
terial most  commonly  used  for 

flat  belts  is  leather.  For  some  kinds  of  work,  however,  belts  woven  from 
cotton  or  other  similar  material  are  used.  When  the  belt  is  to  be 
run  in  a  place  where  there  is  much  moisture,  it  may  be  made  largely 
of  rubber  properly  combined  with  fibrous  material  in  order  to  give 
strength. 


24  ELEMENTS  OF  MECHANISM 

Leather  belts  are  made  by  gluing  or  riveting  together  strips  of  leather 
cut  lengthwise  of  the  hide,  near  the  animal's  back.  If  single  thick- 
nesses of  the  leather  are  fastened  end  to  end,  the  belt  is  known  as  a 
single  belt  and  it  is  usually  about  -f$  in.  thick.  If  two  thicknesses 
of  leather  are  glued  together,  flesh  side  to  flesh  side,  the  belt  is  known 
as  a  double  belt  and  is  from  ^  to  f  in.  thick.  The  manner  of  uniting 
the  ends  of  the  strips  to  form  a  belt,  and  of  fastening  together  the 
ends  of  the  belt  to  make  a  continuous  band  for  running  over  pulleys, 
is  very  important.  A  detailed  discussion  of  these  features  is  not 
necessary,  however,  in  the  present  study  of  the  subject. 

Leather  belts  always  should  be  run  with  the  hair  side  against  the 
pulleys,  if  possible. 

48.  Power  of  Belting.     The  amount  of  power  which  a  given  belt 
can  transmit  depends  upon  its  speed,  its  strength  and   its  ability  to 
adhere  to  the  surface  of  the  pulleys.     The  speed  is  usually  assumed 
to  be  the  same  as  the  surface  speed  of  the  pulleys.     The  strength,  of 
course,  depends  upon  the  width  and  thickness  and  upon  the  nature 
of  the  material  of  which  the  belt  is  made.     The  ability  to  cling  to  the 
pulley  so  as  to  run  with  little  or  no  slipping  depends  upon  the  condi- 
tion of  the  pulley  surfaces  and  of  the  surface  of  the  belt  which  is  in 
contact  with  the  pulleys,  and  upon  the  tightness  with  which  the  belt 
is  stretched  over  the  pulleys.     Since  leather  belts  are  more  common 
and  more  nearly  uniform  in  their  character  than  those  of  other  ma- 
terials, the  discussion  of  power  will  be  confined  to  them. 

49.  Unit  of   Power  —  Horse   Power.     In   order   to   measure  the 
power,  or  the  amount  of  work  done,  by  any  force,  it  is  necessary  to 
have  some  standard  of  measurement.     A  common  unit  for  measur- 
ing work  done  is  that  known  as  the  foot-pound.     A  foot-pound  is  the 
amount  of  work  done  in  raising  a  one-pound  weight  a  distance  of 
one  foot,  or  in  moving  any  number  of  pounds  through  such  a  dis- 
tance that  the  product  of  the  force  exerted  multiplied  by  the  distance 
moved  is  equal  to  one.     For  example,  if  a  12-lb.  weight  is  lifted  one- 
twelfth   of   a   foot,    the   work   done   is    12   Ibs.  X  TV  ft.  =  1    ft.-lb. 
If  the  apparatus  furnishing  the  force  to  raise  this  weight  is  such  that 
it  can  raise  it  in  one  minute,  the  apparatus  is  said  to  be  capable  of 
doing  one  foot-pound  of  work  per  minute,  or  to  have  a  power  of  one 
foot-pound  per  minute. 

For  measuring  large  quantities  of  power  a  larger  unit  is  used, 
known  as  a  horse  power.  One  horse  power  is  equal  33,000  ft. -Ibs.  of 
work  per  minute.  For  example,  an  engine  which  is  capable  of  doing 
one  horse  power  work  is  one  which  can  move  1  Ib.  through  a  distance 
of  33,000  ft.  per  minute,  or  33,000  Ibs.  1  ft.  per  minute,  or  any  num- 


BELTS,   ROPES  AND  CHAINS  25 

her  of  pounds  through  such  a  distance  in  a  minute  that  the  product 
of  the  force  multiplied  by  the  distance  moved  in  a  minute  is  33,000. 

50.  Tension  in  a  Belt.     In  Fig.  29  suppose  the  pulley  A  is  fast  to 
the  shaft  S  and  the  pulley  B  fast  to  the  shaft  Si.    Let  it  be  assumed 
that  when  the  shafts  are  at  rest  a  belt  is  stretched  over  the  pulleys 
as  shown,  the  tightness  with  which  it  is  stretched  being  such  that 
there  is  a  tension  or  pull  in  the  belt  of  a  definite  number  of  pounds. 
This  tension  is  practically  the 

same  at  all  places  in  the  belt 
and  is  called  the  initial  tension. 
Let  this  initial  tension  be  rep- 
resented by  the  letter  TQ. 
Suppose  now  that  some  external 
force  is  applied  to  the  shaft  S  FlG 

causing  it   to   tend   to  turn  in 

the  direction  indicated  by  the  arrow.  This  tendency  to  turn  will 
increase  the  tension  in  the  lower  part  of  the  belt  (say  between  m  and  ri) 
and  decrease  the  tension  in  the  upper  part.  Let  the  new  tension  in  the 
lower  or  tight  side  of  the  belt  be  represented  by  TI  (which  is  greater 
than  To)  and  the  tension  in  the  upper  or  slack  side  by  T2  (which  is  less 
than  To). 

If  the  belt  sticks  to  the  pulley  B  so  that  there  is  no  slipping,  the 
force  TI  tends  to  cause  the  pulley  B  to  turn  as  shown  by  the  full 
arrow  and  the  force  T*  tends  to  cause  B  to  turn  as  shown  by  the 
dotted  arrow.  As  soon  as  T±  becomes  enough  greater  than  T2  to 
overcome  whatever  resistance  the  shaft  Si  offers  to  turning,  the 
pulleys  will  begin  to  turn  in  the  direction  of  the  full  arrow.  The 
unbalanced  force,  then,  which  makes  the  driven  pulley  B  turn  is  the 
difference  between  the  tension  TI  on  the  tight  side  of  the  belt  and  the 
tension  T2  on  the  slack  side  of  the  belt.  This  difference  in  tensions 
is  called  the  effective  pull  of  the  belt  and  is  here  represented  by  the 
letter  E. 

From  the  above  discussion  it  may  be  seen  that  the  following  equa- 
tion holds  true: 

Ti-T2  =  E.  (18) 

51.  To  Find  the  Horse  Power  of  a  Belt.     Since,  as  explained  in 
the  previous  paragraph,  the  effective  pull  is  the  force  in  the  belt  which 
enables  it  to  do  work,  it  follows  that  the  product  of  the  effective  pull 
multiplied  by  the  speed  of  the  belt  in  feet  per  minute  will  give  the 
foot-pounds  of  work  per  minute  that  the  belt  performs,  and  this  divided 
by  33,000  will  give  the  horse  power  which  the  belt  transmits.     If  N  is 


26  ELEMENTS  OF  MECHANISM 

the  r.p.m.  of  S,  and  D  the  diameter  of  pulley  A  (in  feet)  the  following 
equation  expresses  the  horse  power  of  the  belt. 

Belt  speed  in  ft.  per  minute  X  E  _  ^  p  (    . 

33,000 


33,000 

It  is  evident  from  the  above  that  for  a  given  belt  speed  the  greater 
the  difference  between  TI  and  T2  the  more  horse  power  the  belt  trans- 
mits. Any  figures  which  may  be  given  for  the  maximum  allowable 

T 

stress  in  a  belt  and  the  maximum  ratio  -=?  are  necessarily  approximate 

J2 

and  somewhat  a  matter  of  opinion.  For  the  purpose  of  illustrat- 
ing the  method  of  calculating  the  power  which  a  given  belt  might 

T 
be  expected  to  transmit  it  will  be  assumed  that  -^  may  not  exceed  £ 

-/2 

and  that  the  maximum  allowable  tension  per  inch  of  width  is  140 
pounds  for  a  double  leather  belt  and  75  pounds  for  a  single  belt.  If  it 
is  still  further  assumed  that  stresses  due  to  centrifugal  force  may  be 
neglected, 

TI      7 

-?fj-  =  =  and  TI  =  140  Ib.  per  inch  of  width, 

1  2          «S 

whence 

Tl-T2  =  80  Ib. 

=  maximum  effective  pull  per  inch  of  width  of  double  belt. 

Also          ^  =  I  and  TI  =  75, 

1  2          O 

whence 

T1-T2  =  43  Ib.  (nearly) 

=  maximum  effective  pull  per  inch  of  width  of  single  belt. 

Substituting  these  values  in  Eq.  (19), 
Belt  speed  in  ft.  per  minute  X  4$  X  width  of  belt  in  inches 

88,000 

=  H.P.  a  single  belt  will  transmit.     (21) 
Belt  speed  in  ft.  per  minute  X  80  X  width  of  belt  in  inches 

33,000 

=  H.P.  a  double  belt  will  transmit.     (22) 

Corrections  must  be  made  in  equations  (21)  and  (22)  if  the  belt  speed 
is  such  that  centrifugal  force  must  be  taken  into  account.  It  must  also 
be  borne  in  mind  that  the  figures  43  and  80  are  subject  to  modification, 


BELTS,  ROPES  AND  CHAINS  27 

A  simple  and  somewhat  more  conservative  rule  for  estimating  the 
power  of  a  belt  is  known  as  the  millwrights'  rule  and  has  been  deter- 
mined largely  by  experience.  This  rule  is  as  follows: 

A  single  belt  traveling  1000  ft.  per  minute  will  transmit  1  H.P.  per 
inch  of  width  and  a  double  belt  traveling  560  ft.  per  minute  will  transmit 
I  H.P.  per  inch  of  width. 

Whence 

Belt  speed  in  ft.  per  minute  X  width  of  belt  in  inches  _ 
1000 

H.P.  a  single  belt  will  transmit.     (23) 
Belt  speed  in  ft.  per  minute  X  width  of  belt  in  inches  _ 
560 

H.P.  a  double  belt  will  transmit.     (24) 

Example  6.  A  shaft  carrying  a  48-in.  pulley  runs  at  a  speed  of  180  r.p.m.  An 
8-in.  double  belt  runs  over  the  pulley  and  drives  another  shaft.  To  find  the  power 
that  the  belt  can  be  expected  to  transmit  without  excessive  strain. 

Solution  1.    Using  formula  (22), 

Belt  speed  in  feet  =  X  180  =  2262  ft.  per  minute. 

.P.  (nearly). 

Solution  2.     Using  formula  (24) 

Belt  speed  as  in  solution  1  =  2262, 
2262  X  8 


560 


=  32  H.P. 


It  will  be  noticed  that  the  two  solutions  given  above  give  widely  dif- 
ferent answers,  that  from  the  millwrights'  rule  being  nearly  33  per  cent 
less  than  the  other.  For  the  higher  belt  speeds  this  difference  will  not 
be  as  great  if  proper  allowance  is  made  for  centrifugal  force.  Any 
such  solution  for  a  belt  must  be  considered  approximate,  and  merely 
furnishes  a  means  of  estimating  the  horse  power  roughly.  There  is  no 
doubt  that  the  above  belt,  if  in  proper  condition,  would  carry  much 
more  than  even  the  43J  H.P.,  but  the  heavier  the  belt  is  loaded  the 
more  attention  it  will  require  and  the  shorter  will  be  its  life. 

Example  7.  A  shaft  running  200  r.p.m.  is  driven  by  a  single  belt  on  a  24-in. 
pulley.  15  H.P.  is  required.  To  find  a  suitable  width  of  belt  to  use. 

Solution  1.     Using  formula  (21), 


Belt  speed  =          X  200  =  1257  ft.  per  minute. 

1257  X  43  X  width 

33,000          -=15' 

„,.,,,        15  X  33,000 

Width  ==  9  in.  nearly. 


28  ELEMENTS  OF  MECHANISM 

Solution  2.     Using  formula  (23), 

1257  X  width 

1000  ' 

Width  =  y^?  =  12  in.  nearly. 


Here  again  the  millwrights'  rule  shows  a  wider  belt  necessary  for  a 
given  horse  power. 

Example  8.  Two  shafts  A  and  B  are  to  be  connected  by  a  12-in.  double  belt 
carrying  72  H.P.  A  is  the  driving  shaft,  making  240  r.p.m.  B  is  to  run  180  r.p.m. 
To  find  the  size  of  the  pulleys  on  A  and  B}  using  the  millrights'  rule. 

First  find  the  necessary  belt  speed  using  Eq.  (24). 

Belt  speed  X  12  _ 

560  72' 

72  V  ^fif) 
.'.      Belt  speed  =       ^        =  3360  ft.  per  minute. 

Since  B  is  to  turn  180  r.p.m.,  if  x  =  the  diameter  of  the  pulley  on  B,  then, 

TT  x  X  180  =  3360 


or  x  =          -  =  5.94  ft.  =  71.28  in. 

loU  7T 

or,  since  pulleys  of  that  size  would  not  be  made  in  fractional  inches,  a  72-in.  pulley 
would  be  used. 

Pulley  on  A  _  180 

Pulley  on  B  ~  240  ' 

whence  Pulley  on  A  =  72  X  $fft  =  54  in. 

52.  Approximate  Formula  for  Calculating  the  Length  of  Belts.  In 
finding  the  length  of  belt  required  for  a  known  pair  of  pulleys  at 
a  known  distance  apart,  the  most  satisfactory  method,  when  possible, 
is  to  stretch  a  steel  tape  over  the  actual  pulleys  after  they  are  in  posi- 
tion, making  a  reasonable  allowance  (about  1  "  in  every  10  ft.)  for  stretch 
of  the  belt.  Often,  however,  it  is  necessary  to  find  the  belt  length  from 
the  drawings  before  the  pulleys  are  in  place  or  when,  for  some  other  rea- 
son, it  is  not  convenient  actually  to  measure  the  length.  Various  formu- 
las have  been  devised  by  which  the  length  may  be  calculated  when  the 
pulley  diameters  and  distance  between  centers  of  the  shafts  are  known. 
These  formulas,  if  exact,  are  all  more  or  less  complex  and  are,  of  course, 
different  for  crossed  and  for  open  belts.  If  the  distance  between 
shafts  is  large,  the  following  will  give  an  approximate  value  for  the 
length  of  the  belt.  Referring  to  Fig.  31  and  letting  L  represent  the 
length  of  the  belt, 

L  =  ^  +  *>+2C.  (25) 


BELTS,  ROPES  AND  CHAINS 


29 


D,  d  and  C,  must  be  expressed  in  like  linear  units;  if  in  feet,  the  result- 
ing value  of  L  will  be  in  feet;  if  in  inches,  the  value  of  L  will  be  in 
inches. 

In  the  case  of  an  open  belt  where  the  two  pulleys  are  of  the  same 
diameter  the  above  formula  gives  an  exact  answer.  If  the  pulleys  are 
not  of  the  same  diameter,  the  length  of  belt  obtained  by  Eq.  (25)  will 
be  less  than  the  correct  length.  If  the  shafts  are  several  feet  apart 
and  the  difference  in  diameters  of  the  pulleys  is  not  great,  the  percent- 
age error  is  very  small  for  an  open  belt.  With  a  crossed  belt,  pulleys 
of  medium  size  and  the  shafts  several  feet  apart,  the  result  from  the 
use  of  Eq.  (25)  is  considerably  less  than  the  real  length.  This  equation 
is  accurate  enough  to  use  for  an  estimate  of  the  length  of  a  belt. 

53.  Exact  Formulas  for  Length  of  Belt  Connecting  Parallel  Axes. 
While  the  methods  given  in  the  preceding  paragraph  are  sufficient  for 
the  conditions  there  referred  to,  it  is  necessary  in  designing  certain 
pulleys,  known  as  stepped  pulleys  and  cone  pulleys,  to  make  use  of  an 
equation  expressing  exactly,  or  very  nearly  so,  the  belt  length  in  terms 
of  the  diameters  and  the  distance  between  centers  of  the  pulleys.  The 
crossed  belt  and  the  open  belt  must  be  considered  separately. 


FIG.  30 

Crossed  Belts.  Let  D  and  d  (Fig.  30)  be  the  diameters  of  the 
connected  pulleys;  C  the  distance  between  their  axes;  L  the  length  of 
the  belt. 


Then 


L  =  2  (mn  +  no  +  op) 


(26) 


where 


...       at       an  +  bo 

sin  6  =  -T  = ~ 

ab  ao 


30 


ELEMENTS  OF  MECHANISM 


Open  Belts.     Using  the  same  notation  as  for  crossed  belts,  we 
have  (Fig.  31) 

L  =  2  (mn  +  no  +  op) 


(27) 


2<7cos0, 


where          sin  ,  -  «-*  -  »=* 


and  cos  0  = 


FIG.  31 

For  an  open  belt,  6  is  generally  small,  so  that  0  =  sin  0,  very  nearly; 
then 


,  nearly, 


,  nearly. 


If  the  quantity  under  the  radical  sign  is  expanded,  and  all  terms  hav- 
ing a  higher  power  of  C  than  the  square  in  the  denominator  are  neglected, 
since  C  is  always  large  compared  with  (D  —  d), 


(D  - 


or 


L  =     (D  +  d)  +  2  C  + 


4C2 
(D- 


+  1- 


(D  ~ 


8C2 


very  nearly. 


(28) 


54.  Stepped  Pulleys.  Sometimes  it  is  necessary  to  have  such  a 
belt  connection  between  two  shafts  that  the  speed  of  the  driven  shaft 
may  be  changed  readily  while  the  speed  of  the  driving  shaft  remains 
constant.  One  method  of  accomplishing  this  is  the  use  of  a  pair  of 


BELTS,  ROPES  AND  CHAINS 


31 


pulleys  each  of  which  has  several  diameters  as  shown  in  Fig.  32.  Such 
pulleys  are  known  as  stepped  pulleys.  Suppose  that  the  shaft  S,  Fig. 
32,  is  the  driver,  making  N  r.p.m.  When  the  belt  is  in  the  position 
shown  in  full  lines,  the  working  diameter  of  pulley  A  is  A  and  the 
working  diameter  of  pulley  B  is  d\. 
Then  if  ni  represents  the  r.p.m.  of 
Si,  when  the  belt  is  in  this  place, 


Di 


N 


If  the  belt  is  shifted  to  any  other 
position,  as  that  shown  by  dotted 
lines,  Dx  becomes  the  working  diam- 
eter of  the  driving  pulley  and  dx  of 
the  driven  pulley.  If  nx  represents 
the  speed  of  S  for  this  belt  position 

nx  _  Dx 

N~  d  ' 

Therefore,  by  properly  proportioning 
the  diameters  of  the  different  pairs 
of  steps,  it  is  possible  to  get  any 
desired  series  of  speeds  for  the  driven 
shaft. 

In  designing  such  a  pair  of  pulleys 
two  things  must  be  taken  into  ac- 
count. First,  the  ratio  of  the  diam- 
eters of  the  successive  pairs  of  steps 
must  be  such  as  to  give  the  desired 
speed  ratios.  Second,  the  sum  of  the  diameters  of  any  pair  of  steps 
must  be  such  as  to  maintain  the  proper  tightness  of  the  belt  for  all 
positions.  This  second  consideration  makes  the  problem  of  design 
considerably  more  complicated. 

Two  cases  arise:  First,  the  design  of  the  pulleys  for  a  crossed  belt 
and  second,  the  design  for  an  open  belt. 

55.  Stepped  Pulleys  for  Crossed  Belt.  Assuming  that  the  value  of 
DI,  N,  HI,  nx  and  C  are  known  for  the  drive  shown  in  Fig.  32,  and  as- 
suming that  the  belt  is  crossed,  instead  of  open  as  there  shown,  let  it 
be  required  to  find  a  method  for  calculating  Dx  and  dx. 

First  find  di,  which  is  readily  done  from  the  equation 


cT 

n 

m 

1 

j> 

^ 

CQ 

m 

B 


FIG.  32 


N 


32 


ELEMENTS  OF  MECHANISM 


in  which  di  is  the  only  unknown  quantity.     Knowing,  then,  DI  and  dL 
the  value  of  DI  +  di  is  known. 
From  Eq.  (26)  the  length  of  the  belt  to  go  over  the  steps  DI  and  di  is 


(Z)i  +  di)  +  2  C  cos  0i. 


When  the  belt  is  on  the  steps  whose  diameters  are  Dx  and  dx  the 
equation  for  the  length  of  the  belt  is 


2CCOS0S. 


Since  the  same  belt  is  to  be  used  on  both  pairs  of  steps  the  value  of 
these  two  equations  must  be  the  same. 

Therefore, 


+  4)  +  2  C  cos0*. 

Since  C  is  a  constant  and  0  is  dependent  upon  C  and  D  +  d  it  follows 
that  the  above  equation  will  be  satisfied  if 

Ds  +  ds  =  Di  +  di.  (29) 

Therefore  in  designing  a  pair  of  stepped  pulleys  for  a  crossed  belt  the 
sum  of  the  diameters  of  all  pairs  of  steps  must  be  the  same. 
Then  from  the  equation 


and  Eq.  (29) 


_ 
N~  dx 

Dx  +  dx  =  A  +  4. 
Dx  and  dx  may  be  found  by  the  method  of  simultaneous  equations. 


Example  9.    To  find  the  diameters  of  all  the  steps  in  the  pulleys  shown  in  Fig. 
33  if  a  crossed  belt  is  to  be  used. 


First,  find  di  from  the  equation  ^  =  -y-1 


or 

whence 

Therefore, 
From  Eq.  (29) 

and 


192      16 
120  =  &' 


20  R.P.M. 


16  X  120 


D!  +  di  =  16  +  10  =  26  in. 
Z>2  +  d2  =  D!  +  di  =  26  in. 

Dt  =  160 
<k  ~  120 


192  R.P.M,-*, 

160  R.P.M.=na 
80  R.P.M.=na 


FIG.  33 


BELTS,  ROPES  AND  CHAINS  33 

or  Dz  =  f  dz. 

Substituting  this  value  of  Z)2  in  the  preceding  equation, 

f  dz  +  d2  =  26 
or  |rf2  =  26, 

whence  d2  =  —  =  —  =  H-f  in.   =  11.14  in. 

and  D2  =  26  -  11£  =  14f  in.  =  14.86  in. 

Again,  D3  +  dz  =  26  in. 

D3       80 
^  =  I20 

or  Z>3  =  |  </s, 


or  fd3  =  26in., 

whence  d3  =  15f  in.  =  15.6  in. 

and  D3  =  26  -  15f  =  10f  in.  =  10.4  in. 

56.   Stepped  Pulleys  for  Open  Belt.    Referring  still  to  Fig.  32,  if  the 
belt  is  open  its  length  when  on  the  steps  DI  and  di  is,  from  equation  (28), 


and  when  on  steps  Dx  and  dx 

T        T  /r>     i    J\    i    o^    i    (Dx- 
L  =  <£  (Vx  +  ax  )  +  2  C  H  ---  4~ 

Equating  these  two  expressions  gives 

x  m  _i_  ^  \  _i_  (-Pi  —  ^i)2      TT  /n    ,    j  N 
-  (Di  +  rfi)  +  =(DX+  dx) 


nx      DX 
This  may  be  solved  simultaneously  with  ^  =  -^  to  get  the  values  of 

J\        ax 

Dx  and  dx.* 

If  the  shafts  are  several  feet  apart  and  the  range  of  speeds  for  the 
driven  shaft  is  not  excessive  the  diameters  calculated  for  an  open  belt 
differ  only  very  slightly  from  those  for  a  crossed  belt,  and  stepped 
pulleys  designed  for  a  crossed  belt  are  often  used  for  an  open  belt.  If 
the  shafts  are  close  together  and  the  speed  range  is  large  the  crossed 
belt  pulleys  cannot  be  used  for  an  open  belt. 

*  Equation  (30)  may  be  written  in  the  form 

Dx  +  dx  =  DI  +  di  H p: — 79 • 

2  irC 

This  may  be  solved  approximately,  in  connection  with  —  =  -^->  by  substituting  for 

N       dx 

(Dx  —  dx)2  the  value  which  it  would  have  if  the  belt  were  crossed. 


34 


ELEMENTS  OF   MECHANISM 


Example  10.     To  find  the  diameters  of  all  the  steps  in  the  pulleys  shown  in  Fig. 
34,  if  an  open  belt  is  to  be  used.    Shafts  24"  on  centers. 

First  find  d\  from  the  equation 

ni  _  A 


or 


900  _  18 
150  "  di 


whence 


150  X  18 
900 


To  find  D2  and  d2  substitute  in  equation  (30)  the  values  of  A,  di  and  C, 


whence 
and 


(18 


rc2      Z>2         ,          450      Z>2        ~ 
N  =  j2>     whence^  =  ^-  or  Z)2  = 


Substituting  this  value  for  Z>2  and  solving, 

d2  =  5.43  in.      Z>2  =  16.29  in. 

Similarly, 


and 


»  (18  +  3)  +  .  |(Z>S  +  *)  + 

^  =  T'    whence  -^  =  5?  or  4=  2  D., 


c? 


/s 

ISO 


Substituting  and  solving, 

A  =  7.38  in.    and    d3  =  14.76  in. 

The  proportion  chosen  in  the  data  for  Example  10  gives  an  extreme 
case,  and  it  will  be  noticed  that  even  here  the  amount  that  D2  +  dz 

varies  from  DI  +  d\  is  only  about  f  in. 
and  the  variation  of  D»  +  d3  is  a  trifle 
less  than  1-jft-  in.  These  quantities 
are  large  enough  to  affect  the  tightness 
of  the  belt  and  must,  therefore,  be 
taken  into  account.  In  ordinary  cases, 
however,  where  the  distance  between 
centers  is  much  larger  than  in  Example 
10  and  where  the  speed  ratios  are  not 
so  great  the  value  of  Dx  +  dX)  as 
obtained  from  Eq.  (30)  by  the  method 
just  illustrated,  differs  but  very  little 
from  DI  +  di  and  this  difference  can 
usually  be  neglected. 

57.  Equal  Stepped  Pulleys.  It  is 
common  practice,  when  convenient, 
to  design  a  pair  of  stepped  pulleys  in  such  a  way  that  both  pulleys 
have  the  same  dimensions  and  can,  therefore,  be  cast  from  the  same 


FIG.  34 


BELTS,  ROPES  AND  CHAINS 


35 


pattern.     This  condition  imposes  certain  restrictions  on  the  speed  ratios 
as  may  be  seen  from  the  following: 

Referring  to  Fig.  35  if  the  pulleys  are  alike, 


As  in  previous  discussions, 


and 


but 


N~  d, 
D5      d, 


Therefore, 

In  a  similar  manner, 
and 


(31) 


N 


=  ns. 


ft 


ft 


ft 


, 


N-R.P.M, 


ft 


</20 


A 


cfc 


FIG.  35 


FIG.  36 


That  is :  When  equal  stepped  pulleys  are  used  the  speeds  of  the  driven  shaft 
must  be  so  chosen  that  the  speed  of  the  driving  shaft  is  a  mean  proportional 
between  the  speeds  of  the  driven  shaft  for  belt  positions  symmetrically 
either  side  of  the  middle  step. 

Example  11.    A  pair  of  equal  three-stepped  pulleys,  Fig.  36,  are  to  carry  a 
belt  to  connect  two  shafts.    The  driving  shaft  makes  120  r.p.m.,  and  the  lowest 


36 


ELEMENTS   OF  MECHANISM 


speed  of  the  driven  shaft  is  60  r.p.m.     To  find  the  other  two  speeds  of  the  driven 
shaft. 

r*     N 

N      nz 

ni_  _  120 

120  "  60 ' 

Therefore  HI  =  240, 

n2  =  N  =  120. 

If  the  step  diameters  are  to  be  calculated,  it  will  be  done  by  the  methods  explained 
in  §  55  or  §  56  according  as  the  belt  is  crossed  or  open. 


Driver. 


FIG.  37 


58.  Speed  Cones.  In  some  cases  instead  of  stepped  pulleys, 
pulleys  which  are  approximately  frusta  of  cones  are  used,  as  shown  in 
Fig.  37.  Here  the  working  diameters  of  the  pulleys,  as  Dx  and  dx  for 
any  belt  position,  are  measured  at  the  middle  of  the  belt.  To  design 


BELTS,  ROPES  AND  CHAINS 


37 


such  a  pair  of  pulleys  a  series  of  diameters  Di,  A>,  D3,  etc.  (Fig.  38), 
may  be  calculated  in  the  same  way  as  steps  and  plotted  at  equal  dis- 
tances (a)  apart,  then  a  smooth  line  drawn  through  their  ends,  as 
shown.  The  length  (a)  does  not  affect  the  problem  except  as  it  makes 
the  cone  longer  or  shorter.  The  contours  may  be  straight  lines  as  in 
Fig.  38,  giving  cones,  or  curves  as  in  Fig.  39,  giving  conoids. 

When  cones  are  used,  a  shipper  must  guide  each  part  of  the  belt  just 
at  the  point  where  it  runs  on  to  the  pulley  (see  Fig.  37) ;  otherwise 
the  belt  will  tend  to  climb  toward  the  large  end  of  each  pulley. 
Both  shippers  must  be  moved  simultaneously  when  the  belt  is 
shifted. 


.-1. 


io    o  _yv 

^^A^ 

^U  ^    .J 

I 

FIG.  38 


FIG. 


59.    Belt   Connections  between   Shafts   which  are  not  Parallel. 

Non-parallel  shafts  may  be  connected  by  a  flat  belt  with  satisfactory 
results,  provided  the  pulleys  are  so  located  as  to  conform  to  a  funda- 
mental principle  which  governs  the  running  of  all  belts,  namely:  The 
point  where  the  pitch  line  of  the  belt  leaves  a  pulley  must  lie  in  a  plane 
passing  through  the  center  of  the  pulley  toward  which  the  belt  runs. 
In  other  words,  a  plane  through  the  center  of  the  receiving  pulley,  per- 
pendicular to  the  pulley  axis,  must,  if  produced,  include  the  delivering 
point  of  the  pulley  from  which  the  belt  is  running.  This  may-  be  seen 
by  a  reference  to  Fig.  40.  In  this  case  the  shafts  S  and  T  are  intended 
to  turn  in  the  directions  indicated  by  the  arrows. '  Considering  Eleva- 
tion A,  the  pitch  line  of  the  belt  leaves  the  pulley  M  at  the  point  a.  If 
the  pulley  N  is  in  such  a  position  on  the  shaft  T  that  a  plane  through 
the  middle  of  its  face  contains  the  point  a,  the  belt  will  run  properly 
on  to  pulley  N.  XX  is  the  trace  of  this  plane  and  evidently  contains 
point  a.  Similarly,  in  Elevation  B,  the  pitch  line  of  the  belt  leaves  the 
pulley  N  at  61  and  M  is  so  located  on  shaft  S  that  a  plane  YY  through 
the  middle  of  its  face  contains  61. 

Fig.  41  shows  the  proper  relative  position  of  the  pulleys  if  the  direc- 
tion of  turning  of  shaft  S  is  the  reverse  of  that  in  Fig.  40.  Other 
changes  in  the  directions  of  rotation  of  either  pulley  would  necessitate 
corresponding  changes  in  the  relative  positions. 


38 


ELEMENTS  OF  MECHANISM 


Both  Figs.  40  and  41  show  the  pulleys  at  90°  with  each  other.  The 
belt  would  run  equally  well  if  the  pulleys  were  turned  at  any  angle  about 
XX  as  an  axis. 


Elevation  A 


Elevation  B 


FIG.  40 


60.  Quarter-turn  Belt.  A  belt  which  connects  two  non-intersect- 
ing shafts  at  right  angles  with  each  other,  similar  to  those  in  Figs.  40 
and  41,  is  called  a  quarter-turn  belt.  Emphasis  should  be  laid  on  the 
fact  that,  for  any  given  setting  of  the  pulleys,  the  shafts  must  always 


BELTS,  ROPES  AND  CHAINS 


39 


turn  in  the  direction  in  which  they  were  designed  to  turn.  If  the 
direction  of  rotation  is  changed  without  resetting  the  pulleys,  the  belt 
will  immediately  leave  the  pulleys.  For  this  reason  simple  quarter- 


FIG.  41 

turn  belts  like  those  illustrated  above  are  likely  to  give  trouble  if 
used  in  places  where  there  is  possibility  of  the  shafting  turning  back- 
wards even  a  small  fraction  of  a  turn.  If  this  should  happen  to  a 


40  ELEMENTS  OF  MECHANISM 

small  belt,  it  could  easily  be  replaced  on  the  pulleys;  in  the  case  of  a 
large  belt,  however,  the  replacing  would  be  more  difficult. 

61.  Reversible   Direction  Belt  Connection  between  Non-parallel 
Shafts  —  Guide    Pulleys.     If    the    connection    between    two    non- 
parallel  shafts  is  to  be  such  that  the  shafting  may  run  in  either  direc- 
tion and  still  have  the  pulleys  deliver  the  belt  properly,  in  accordance 
with  the  fundamental  law  already  explained,  it  is  necessary  to  make 
use  of  intermediate  pulleys  to  guide  the  belt  into  the  proper  plane. 
Such  pulleys  are  called  guide  pulleys. 

62.  Examples  of  Belt  Drives  —  Method  of  Laying  Out.     The  fol- 
lowing examples  will  illustrate  a  few  of  the  types  of  belt  drives  which 

may  occur  and  will  give  some 
idea  of  the  method  of  pro- 
cedure in  designing  such  drives. 
Some  of  these  examples  are 
chosen  from  existing  drives; 
others  have  been  modified  in 
order  to  illustrate  the  principles 
more  clearly. 

Example  12.     Given  two  shafts  A 
and  B  located  as  shown  in  Fig.  42. 
Shaft  A,  carrying  a  52-in.  pulley,  is 
SB  to  drive  a  60-in.  pulley  on  shaft  B  by 

— -  =3 —     means  of  a  12-in.  double  belt.     Two 

V  5  guide  pulleys  30  ins.  diameter  in  line 

Front  Elevation  Right  Elevation       with  each  other  are  to  be  located  on 

FIG.  42  two  horizontal  shafts  so  that  the 

direction  of  rotation  may  be  reversed 

without  the  belt  running  off.  When  turning  in  the  direction  indicated  by  the 
arrows,  the  tight  side  of  the  belt  is  to  run  direct  from  the  driven  to  the  driving  pulley 
in  a  vertical  line,  the  loose  side  returning  around  the  guide  pulleys.  The  guide  pulley 
which  receives  the  belt  from  the  upper  main  pulley  is  to  be  on  a  shaft  whose  center 
is  9  ft.  below  the  center  of  A }  the  other  guide  pulley  is  to  be  on  a  shaft  whose  center 
is  2  ft.  6  in.  below  the  center  of  A.  Main  pulleys  16-in.  face,  guide  pulleys  will  be 
drawn  the  same  width  as  the  belt  although  in  practice  they  would  be  wider. 

To  draw  two  elevations  and  a  plan. 

Solution.  Referring  to  Fig.  43,  first  draw  the  center  line  YYi  which  is  the  center 
line  through  the  shaft  B.  At  a  distance  of  14  ft.  above  YYi  draw  XXi  as  the  center 
line  of  shaft  A.  At  any  convenient  place  near  the  left  end  of  YYi  choose  the  point 
BL  as  the  center  point  of  shaft  B  for  the  left  elevation.  With  BL  as  a  center  draw 
a  circle  60-ins.  diameter  which  will  be  the  left  elevation  of  the  pulley  on  B.  Next, 
draw  the  vertical  line  TTi  tangent  to  this  pulley  on  the  side  which  is  moving  up- 
ward. TTi  is  the  pitch  line  of  the  tight  part  of  the  belt  and  must  be  contained  in 
the  center  plane  of  the  pulley  on  A.  With  TTi  and  XXi  as  center  lines,  draw  the 
rectangle  1-2-3-4  of  length  equal  to  the  diameter  of  the  pulley  on  A  and  width 
equal  to  the  width  of  face  of  the  same  pulley.  This  rectangle  is  the  side  view  or 


BELTS,   ROPES  AND  CHAINS 


41 


I 


FIG.  43 


ELEMENTS  OF  MECHANISM 


left  elevation  of  the  upper  or  driving  pulley.  Next,  choose  a  point  AP  near  the  right 
end  of  XXi  and  draw  the  end  view  or  right  elevation  of  this  pulley.  The  vertical 
line  TIP  Tp  drawn  tangent  to  the  upward  moving  or  driving  side  of  this  pulley  will 
be  the  right  elevation  of  the  driving  or  tight  part  of  the  belt  and  must  be  contained 
in  the  center  plane  of  the  driven  pulley,  since  the  direction  of  rotation  is  to  be  re- 
versible. Therefore,  the  lines  TiP  TP  and  YYl  are  the  center  lines  for  the  rectangle 
5-6-7-8  which  is  the  right  elevation  of  the  driven  pulley,  TTi  and  TP  TIP  are  the 
two  views  of  the  line  of  intersection  of  the  center  planes  of  the  pulleys. 

The  position  of  the  two  main  pulleys  with  respect  to  each  other  has  now  been 
determined  and  the  two  elevations  drawn.  Their  plan  must  next  be  drawn.  This 
may  be  placed  above  either  elevation,  and  is  here  placed  above  the  left  elevation. 
Looking  down  on  the  pulleys,  both  will  appear  as  rectangles.  The  center  line 
XhXih  may  be  drawn  at  any  convenient  distance  above  XXi  and  is  the  horizontal 
projection,  or  plan  view,  of  the  center  line  of  the  shaft  A.  The  pulley  on  this  center 
line  can  be  projected  directly  up  from  the  rectangle  1-2-3-4 Jand  will,  of  course,  have 
the  same  dimensions.  The  center  line  MN  of  the  shaft  B  will  be  vertically  above 
BL  and  the  rectangle  which  forms  the  plan  view  of  the  pulley  on  B  will  be  located 
on  this  center  line  with  its  middle  line  passing  through  the  front  end  of  the  plan  of 
the  other  pulley.  In  other  words,  the  plan  view  of  these  two  pulleys  is  obtained  by 
projecting  from  the  two  elevations  in  accordance  with  the  ordinary  principles  of 
projections. 

To  draw  the  guide  pulleys,  first  locate  them  in  the  plan.  Their  center  plane  will 
contain  the  line  VW  and  one  will  have  its  contour  passing  through  point  V  in  plan 

while  the  contour  of  the  other  will  pass  through 
W.  To  draw  the  elevations,  first  draw  the  center 
lines  at  the  specified  distances  below  shaft  A  and 
then  draw  the  ellipses  which  represent  the  pulleys 
by  projecting  from  the  plan. 

Example  13.  Shaft  S  (Fig.  44)  drives  shaft 
T  by  means  of  an  8-in.  double  belt.  Both  main 
pulleys  36  in.  diameter  located  as  shown.  The 
usual  direction  of  rotation  to  be  as  indicated  by 
the  arrows  but  the  arrangement  to  be  such  that 

Y  \*X  '  4  1 1 1  ,i?         the  directions  may  be  reversed.     Two   15-in. 

\T   ~)  guide  pulleys  are  to  be  placed  on  a  vertical  shaft 

to  carry  the  belt  between  the  two  main  pulleys. 
All  pulleys  9  ins.  face. 

To  draw  the  drive,  making  the  elevations 
and  a  plan. 

Solution.  (See  Fig.  45.)  Draw  the  three  views  of  the  main  shafts  and  pulleys, 
the  plan  and  front  elevation  being  the  same  as  shown  in  Fig.  44,  and  the  right  eleva- 
tion being  constructed  from  these  in  accordance  with  the  usual  principles  of  projec- 
tion. The  left  elevation  might  have  been  made  instead  of  the  right. 

The  position  of  the  guide  pulley  shaft  can  best  be  determined  from  the  plan. 
The  planes  of  the  pulleys  A  and  B  intersect  in  a  line  which,  in  plan,  is  projected  as 
the  point  P  and  in  the  two  elevations  as  the  lines  XY  and  XiYi,  respectively.  Since 
the  upper  guide  pulley  is  to  deliver  the  belt  to  pulley  B,  it  must  be  tangent  to  the 
line  PM ,  and  since,  if  the  direction  of  rotation  is  reversed,  C  must  be  able  to  deliver 
the  belt  to  pulley  A  it  must  be  tangent  to  the  line  PN.  The  same  reasoning  will 
apply  to  the  lower  guide  pulley.  The  center  of  the  guide  pulley  shaft  will,  there- 


P/an 


BELTS,  ROPES  AND  CHAINS 


43 


fore,  be  at  a  point  which  is  distant  from  PM  and  PN  an  amount  equal  to  the  radius 
of  the  guide  pulleys.  Since  the  pulleys  A  and  B  are  of  the  same  diameter  and  their 
axes  on  the  same  level,  the  guide  pulley  C  will  appear  in  the  elevations  with  its 
center  plane  tangent  to  the  tops  of  A  and  B,  and  D  will  have  its  center  plane  tan- 
gent to  the  bottoms  of  A  and  B.  With  this  arrangement  it  is  possible  for  either  of 


Front  Elevation 


Right  Elevation 


FIG.  45 


the  main  pulleys  to  deliver  the  belt  into  the  plane  of  either  guide  pulley,  and  either 
guide  pulley  may  deliver  to  either  main  pulley. 

It  should  be  noticed  that  a  drive  like  this,  with  both  guides  on  the  same  vertical 
shaft,  can  be  reversible  in  direction  only  when  the  main  pulleys  are  of  the  same 
diameter.  The  next  two  examples  show  the  construction  when  the  main  pulleys 
are  of  different  diameters. 

Example  14.  Referring  again  to  Fig.  44,  assume  the  same  conditions  as  for 
Example  13,  except  that  the  main  pulleys  are  of  different  diameters.  Suppose  the 
pulley  on  T  is  48-in.  diameter  and  that  on  S  36-ins.  diameter.  The  direction  of  rota- 
tion not  to  be  capable  of  being  reversed. 

Solution.  See  Fig.  46.  The  three  views  of  the  main  shafts  and  pulleys  are  drawn 
as  in  Example  13.  The  center  of  the  guide  pulley  shaft  is  located  in  the  plan  at 
such  a  point  that  the  pulley  circumference  will  be  tangent  to  the  lines  PM  and  PN 
as  in  Fig.  45.  The  position  of  the  guide  pulley  C  on  this  shaft  is  determined  in  the 
front  elevation,  it  being  at  such  a  height  that  its  center  plane  will  be  tangent  to  the 


44 


ELEMENTS  OF  MECHANISM 


top  surface  (that  is,  contain  the  point  of  delivery  K)  of  the  pulley  A  which  delivers 
the  belt  to  C.  Similarly,  in  the  right  elevation,  the  position  of  the  guide  pulley  D 
is  such  that  its  center  plane  will  be  tangent  to  the  lower  surface  (that  is,  will  contain 
the  point  of  delivery  R)  of  the  pulley  B. 


Front  Elevation 


Right  Elevation 


FIG.  46 


Example  16.  With  the  data  the  same  as  for  Example  14,  suppose  it  is  required 
so  to  arrange  the  guide  pulleys  that  the  direction  of  rotation  may  be  reversed.  (They 
cannot  in  this  case  be  on  the  same  vertical  shaft.) 

Solution.  See  Fig.  47.  After  having  drawn  the  three  views  of  the  main  shafts 
and  pulleys,  the  problem  becomes  one  of  so  placing  the  guide  pulleys  that  they  will 
conduct  the  belt  in  either  direction.  There  are  a  great  many  possible  solutions  of 
this  problem,  but  that  shown  in  Fig.  47  is  the  simplest. 

In  the  front  elevation  the  points  a  and  b  are  the  center  points  of  the  upper  and 
lower  contour  elements  of  the  pulley  B.  From  a  and  6  draw  lines  ae  and  bf  tangent 
to  pulley  A.  The  center  planes  of  the  guide  pulley  C  must  contain  the  line  ae  and 
the  center  plane  of  the  guide  pulley  D  must  contain  the  line  bf.  C  will  appear  in 
this  view,  therefore,  as  a  rectangle  with  one  end  passing  through  a  and  D  will  appear 
as  a  rectangle  with  one  end  passing  through  6.  In  the  other  views  the  edges  of  the 
guide  pulleys  will  appear  as  ellipses,  as  shown. 

Example  16.  The  shaft  S,  Fig.  48,  is  to  drive  the  shaft  T  by  means  of  an  8-in. 
belt  running  on  pulleys  A  and  B.  One  of  the  columns  of  the  building  makes  it  im- 
possible to  continue  the  shaft  T  far  enough  to  place  the  pulley  B  in  the  proper  posi- 


BELTS,   ROPES  AND  CHAINS 


45 


tion  relative  to  A  to  permit  the  use  of  a  direct  quarter-turn  drive.  Furthermore  the 
vertical  distance  between  the  two  shafts  is  too  small  to  make  such  a  drive  practicable 
even  if  the  column  did  not  interfere.  It  is,  therefore,  necessary  to  employ  guide 
pulleys  to  conduct  the  belt  from  B  to  A  and  from  A  back  to  B.  The  relative  direc- 


FRONT  ELEVATION  '    Y 

FIG.  47 


,      RIGHT  ELEVATION 


tions  of  rotation  are  to  be  as  shown  and  the  guide  pulleys  are  to  be  so  located  that  the 
directions  may  be  reversed.  18-in.  guide  pulleys  will  be  used. 

Solution.  See  Fig.  49.  The  positions  of  the  guide  pulleys  are  determined  from 
the  left  elevation.  From  the  center  points  a  and  b  of  the  upper  and  lower  contour 
lines  of  pulley  A,  lines  ae  and  bf  are  drawn  tangent  to  the  pulley  B.  The  center 
plane  of  one  guide  pulley  C  must  contain  the  line  ae  and  the  pulley  will  appear  in 
this  view  as  a  rectangle  with  one  end  passing  through  a.  The  center  plane  of  the 
other  guide  pulley  D  must  contain  the  line  bf  and  the  pulley  will  appear  as  a  rectangle 
with  one  end  passing  through  6.  The  guide  pulleys  will  appear  in  the  front  elevation 
and  in  the  plan  with  their  edges  ellipses,  as  shown.  The  two  guide  pulleys  so  nearly 
coincide  in  the  plan  that  the  lower  one  was  omitted  in  the  drawing. 

It  will  be  noticed  that  in  all  the  preceding  examples  the  same  surface  of  the  belt 
comes  in  contact  with  the  main  pulleys  at  all  times.  This  is  an  important  condi- 
tion from  a  practical  standpoint.  Whenever  practicable  the  same  surface  should 
run  against  the  guide  pulleys  also. 

Example  17.  Given  two  shafts  at  right  angles,  located  as  shown  in  Fig.  50. 
Shaft  A  carries  a  52-in.  pulley  which  drives  a  60-in.  pulley  on  shaft  B  by  means  of 


46 


ELEMENTS  OF  MECHANISM 


Left  Elevation  Front  Elevation 

FIG.  48 


LEFT 
ELEVATION 


FRONT  ELEVATION 


FIG.  49 


BELTS,   ROPES  AND  CHAINS 


47 


a  double  belt  12  ins.  wide.     The  ordinary  direction  of  rotation  is  as  shown  by  the 

arrows.     One  guide  pulley  30  ins.  diameter  is  to  be  so  located  that  the  direction  of 

rotation  may  be  reversed  without  the 

belt  running  off.     When  turning  in  the 

direction  shown,  the  tight  side  of  the 

belt  is  to  run  direct  from  driven  to 

driving  pulley  in  a  vertical  line,  the 

loose  side  returning  around  the  guide 

pulley. 

The  main  pulleys  are  14  ins.  wide. 
Two  elevations  and  a  plan  are  to  be 
drawn. 

Solution.  It  will  be  noticed  that, 
except  for  the  guide  pulley,  this  problem 
is  the  same  as  Example  12  and  the 
method  of  drawing  the  three  views  of 
the  main  pulleys  is  exactly  as  described 
for  that  case. 

To  draw  the  guide  pulley  proceed  as 
follows :  (See  Fig.  5 1 ) .  The  distance  of 


,B 


Front  Elevation          Right  Elevation 
FIG.  50 


this  guide  from  either  one  of  the  main  pulleys  would  be  governed  somewhat  by  con- 
venience in  actually  setting  up  the  bearings  to  support  it,  and  partly  also  by  the 
relative  sizes  of  the  main  pulleys.  It  is  desirable  so  to  locate  it  as  to  give  the  least 
possible  abruptness  to  the  bend  in  the  belt.  In  this  case  there  has  been  selected  a 
point  C  in  the  line  of  intersection  of  the  two  main  pulley  planes  which  is  6  ft.  6  ins. 
below  the  axis  of  the  upper  shaft.  This  point  will  be  at  CL  and  Cp  in  the  two  eleva- 
tions. From  CL  draw  a  line  tangent  to  the  lower  pulley  at  DL  and  project  across, 
getting  the  other  view  of  this  point  at  Dp.  In  a  similar  way  draw  a  line  from  Cp 
tangent  to  the  upper  pulley  at  EP  and  project  across  to  find  EL.  We  now  have  the 
two  projections  of  two  lines  CD  and  CE  drawn  from  a  point  in  the  intersection  of 
the  pulley  planes  tangent  to  the  two  pulleys,  and  the  guide  pulley  must  be  set  in 
such  a  position  that  its  center  plane  will  contain  these  two  lines.  The  problem  then 
is  to  draw  the  projections  of  the  guide  pulley  when  so  set.  Either  elevation  may 
be  drawn  first;  let  us  start  with  the  left  elevation.  Here  CLDL  shows  in  its  true 
length  and  the  line  CD  itself  may  be  considered  as  lying  in  the  plane  of  the  paper. 
The  line  CE  has  one  end  CL  in  the  plane  of  the  paper  while  the  line  itself  really 
slants  down  below  the  paper.  The  true  size  of  the  angle  which  it  makes  with  the 
plane  of  the  paper  is  equal  to  the  angle  EpCpFp.  Dz.Cz,  is  the  trace,  or  line  of  in- 
tersection with  the  paper,  of  the  plane  containing  CD  and  CE. 

Now  swing  this  plane  up  into  the  paper  to  get  the  true  angle  between  CD  and 
CE.  To  do  this  produce  Dz,Cz,  to  the  left  and  through  EL  draw  a  line  perpendicular 
to  DL.CL.  From  CL  with  a  radius  equal  to  the  true  length  of  CE  (that  is,  with  radius 
CpEp)  cut  this  perpendicular  at  E\  and  join  E\  to  CL.  The  angle  EiCLDL  is  the  true 
angle  between  the  lines  EC  and  DC.  Set  the  compasses  with  a  radius  equal  to  the 
radius  of  the  guide  pulley  and  find  by  trial  the  center  Oi  about  which  a  circle  of 
this  radius  may  be  drawn  tangent  to  CiM\  and  CLDL.  This  point  shows  the  real 
position  of  the  center  of  the  middle  circle  of  the  guide  pulley  relative  to  the  lines 
CE  and  CD.  The  next  step  is  to  revolve  Oi  back  to  find  its  projection  relative  to 
CiiEL  and  CLDL.  To  do  this  draw  a  line  through  E\  and  0\  meeting  CLDL  at  H\\ 
then  join  Hi  to  EL.  Through  Oi  draw  a  line  perpendicular  to  Dz.Cz,  meeting  H\EL 
at  OL  and  CLDL  at  JL.  Then  OL  is  the  projection  of  the  center  point  of  the  pulley. 


48 


ELEMENTS  OF  MECHANISM 


w, 


^ 


*r- 


FIG.  51 


BELTS,   ROPES  AND  CHAINS 


49 


To  draw  the  projections  of  the  edges  of  the  pulley  select  any  point  as  «72  in  the  line 

CLDL,  draw  a  line  through  OL  parallel  to  Cz,Z)z,  and  from  J2  with  a  radius  equal  to 

the  radius  of  the  guide  pulley  cut  this  line  at  02.     Draw  a  line  through  J2  and  02, 

also  a  line  VzWz  through  02  perpendicular  to  J202.     On  these  two  lines  as  center 

lines  construct  the  rectangle  W2n2r2s2  of  length  equal  to  the  diameter  of  the  guide 

pulley  and  width  equal  to  the  width  of  face  of  the  same.     This  rectangle  is  the  side 

view  of  the  guide  pulley  when  revolved  up  into 

the  plane  of  the  paper,  the  line  VzWz  being  the 

center  line  of  its  shaft.     The  projection  of  the 

pulley  consists  of  two  equal  ellipses  with  major 

axes  equal  to  the  diameter  of  the  guide  pulley 

and  minor  axes  found  by  projecting  the  points 

m2n2  and  s2r2  on  to  the  line  OL/Z,,  as  shown.     If  a 

definite  length  VzWz  is  chosen  for  the  shaft,  the 

projections  of  the  ends  of  the  shaft  are  found  at 

VL  and  WL.     The  right  elevation  of  the  guide 

pulley  and  the  axis  of  its  shaft  are  found  by  a 

method  exactly  similar  to  that  just  described. 

The  plan  view  of  the  guide  pulley  is  found  by 
first  finding  the  plan  projections  Vh  and  W*  of 
the  two  ends  of  the  shaft  by  projecting  up  from 
the  two  elevations,  then  revolving  this  line  VhWh 
over  until  it  comes  into  the  plane  of  the  paper, 
as  shown  at  FsTFs.  On  this  line  is  drawn  the 
rectangle  tsx3ysZ3,  representing  the  pulley,  and  the 
axes  of  the  ellipses  which  constitute  the  plan  view 
of  the  pulley  are  found  by  projecting  the  rectan- 
gle tzXsy&z  on  to  VhWh,  as  shown. 


FIG.  52 


_ 

\ 

r. 

\i 

_i  

3. 

(L 

a^ 

J 

63.  Crowning  of  Pulleys.  If  a  belt 
is  led  upon  a  revolving  conical  pulley,  it 
will  tend  to  lie  flat  upon  the  conical 
surface,  and,  on  account  of  its  lateral 
stiffness,  will  assume  the  position  shown 
in  Fig.  52.  If  the  belt  travels  in  the  direction  of  the  arrow,  the  point  a 
will,  on  account  of  the  pull  on  the  belt,  tend  to  adhere  to  the  cone  and 
will  be  carried  to  6,  a  point  nearer  the  base  of  the  cone  than  that  pre- 
viously occupied  by  the  edge  of  the  belt  :  the  belt  would  then  occupy  the 
position  shown  by  the  dotted  lines.  Now  if  a  pulley  is  made  up  of  two 
equal  cones  placed  base  to  base,  the  belt  will  tend  to  climb  both,  and 
would  thus  run  with  its  center  line  on  the  ridge  formed  by  the  union  of 
the  two  cones.  In  practice  pulley  rims  are  made  slightly  crowning, 
except  in  cases  where  the  belt  must  occupy  different  parts  of  the  same 
pulley.*  In  Fig.  52  two  common  forms  of  rim  sections  are  shown  at  C 
and  D;  that  shown  at  C  is  most  commonly  met  with,  as  it  is  the  easier 


*  The  amount  of  crowning  varies  from  about 
on  a  pulley  30"  wide. 


a  pulley  6"  wide  to  about 


50 


ELEMENTS  OF   MECHANISM 


to  construct.  When  pulleys  are  located  on  shafts  which  are  slightly 
out  of  parallel,  the  belt  will  generally  work  toward  the  edges  of  the 
pulleys  which  are  nearer  together.  The  reason  for  this  may  be  seen 
from  Fig.  53.  The  pitch  line  of  the  belt  leaves  pulley  A  at  point  a. 
In  order  to  contain  this  point  the  center  plane  of  pulley  B  would  have 
to  coincide  with  XXi.  That  is,  the  belt  is  delivered  from  A  into  the 
plane  XXi.  Similarly,  the  belt  is  delivered  from  6,  on  the  under  side 
of  pulley  B,  into  the  plane  YiY.  The  result  of  this  action  is  that 
the  belt  works  toward  the  left  and  tends  to  leave  the  pulleys. 

64.  Tight  and  Loose  Pulleys  are  used  for  throwing  machinery  into 
and  out  of  gear.  They  consist  of  two  pulleys  placed  side  by  side  upon 
the  driven  shaft  CD  (Fig.  54);  A,  the  tight  pulley,  is  keyed  to  the 
shaft;  while  B,  the  loose  pulley,  turns  loose  upon  the  shaft  and  is  kept 
in  place  by  the  hub  of  the  tight  pulley  and  a  collar.  The  driving  shaft 
carries  a  pulley  G,  whose  width  is  the  same  as  that  of  A  and  B  put 
together,  or  twice  that  of  A.  The  belt,  when  in  motion,  can  be 
moved  by  means  of  a  shipper  that  guides  its  advancing  side,  either 

on  to  the  tight  or  the  loose  pulley.  The 
pulley  G  (Fig.  54)  has  a  flat  face,  because 
the  belt  must  occupy  different  positions 
upon  it,  while  A  and  B  have  crowning 
faces,  which  will  allow  the  shifting  of  the 
belt  and  will  retain  it  in  position  when 
shifted  upon  them. 


C-e}- 

A 


FIG.  53 


FIG.  54 


65.  Ropes  and  Cords.  Power  is  often  transmitted  by  means 
of  ropes  running  over  pulleys,  called  sheaves,  having  grooved 
surfaces.  For  large  amounts  of  power  inside  of  buildings  the 
ropes  are  made  of  hemp  or  similar  material.  For  long  dist- 
ance drives  and  drives  which  are  exposed  to  the  weather  wire 


BELTS,  ROPES  AND  CHAINS 


51 


ropes  are  used.     For  small  amounts  of  power  on  machines,  cords  of 
cotton  are  common. 

66.   Systems  of  Driving  with  Hemp  Rope.     There  are  two  distinct 
systems  of  rope  driving,  each  of  which  has  its  advantages.     One  is 


FIG.  55 


the  Multiple  Rope  or  English  System.  This  is  the  simpler  of  the 
two  and  consists  of  independent  ropes  running  side  by  side  in  grooves 
on  the  pulleys.  A  large  drive  using  this  system  is  shown  in  Fig.  55. 


FIG.  56 


The  other  system  is  the  Continuous  or  American  System,  shown 
in  Figs.  56  and  57.  One  rope  is  wound  around  the  driving  and  driven 
pulleys  several  times,  and  conducted  back  from  the  last  groove  of  one 


52 


ELEMENTS  OF  MECHANISM 


-Driving 


Driven\ 


FIG.  57 


FIG.  58 


Driven 
FIG.  59  FIG.  60 


BELTS,   ROPES  AND  CHAINS 


53 


pulley  to  the  first  groove  of  the  other  pulley  by  means  of  one  or  more 
intermediate  pulleys  which  also  serve  the  purpose  of  maintaining  a 
uniform  tension  throughout  the  entire  rope.  The  slack  should  be  taken 
up  on  the  loose  side  just  off  the  driving  sheave.  There  are  two  ways 


Pitch 
Line 


FIG.  61 


FIG.  62 


of  accomplishing  this.  First  (see  Fig.  56),  the  rope  is  conducted  from 
an  outside  groove  of  the  driver  to  the  tension  sheave  and  after  passing 
around  it  is  returned  to  the  opposite  outside  groove  of  the  driven  sheave. 
Second  (see  Fig.  57),  where  it  is  inconvenient  to  take  the  slack  directly 
from  the  driver  the  rope  is  passed  around  a  loose  sheave  on  the  driven 
shaft,  thence  over  the  tension  sheave,  and  is  returned 
to  the  first  groove  in  the  driven  sheave.  Figs.  58,  59 
and  60  show  further  examples. 

67.  Grooves  for  Hemp  Rope.     The  shape  and 
proportions  of  the  grooves  used  on  many  pulleys  for 
hemp  rope  depend  somewhat  upon  the  system  used. 
Figs.  61  and  62  show  two  forms  much  used.     Fig.  63 
illustrates  the  groove  used  on  idle  wheels. 

It  will  be  noticed  that  the  rope  wedges  into  the 
grooves  on  the  driving  and  driven  pulleys,  while  on  the 
loose  or  idle  pulleys  it  rides  on  the  bottom  of  the  groove. 

68.  Small  Cords  are  often  used  to  connect  non- 
parallel  axes,  and  very  often  the  directional  relation 
of  these  axes  must  vary.     The  most  common  example 

is  found  in  spinning  frames 
and  mules,  where  the  spin- 
dles are  driven  by  cords 
from  a  long,  cylindrical 
drum,  whose  axis  is  at  right 
angles  to  the  axes  of  the 
spindles.  In  such  cases,  the  common  perpendicular  to  the  two  axes  must 
be  contained  in  the  planes  of  the  connected  pulleys;  both  pulleys  may 
be  grooved,  or  one  may  be  cylindrical,  as  in  the  example  given  above. 
Fig.  64  shows  two  grooved  pulleys,  whose  axes  are  at  right  angles  to 
each  other,  connected  by  a  cord  which  can  run  in  either  direction,  pro- 


FIG.  63 


FIG.  64 


54 


ELEMENTS  OF  MECHANISM 


vided  the  groove  is  deep  enough.  To  determine  whether  a  groove  has 
sufficient  depth  in  any  case,  the  following  construction  (Fig.  65)  may  be 
used.  Let  AB  and  A\Bi  be  the  projections  of  the  approaching  side  of 
the  cord;  pass  a  plane  through  AB  parallel  to  the  axis  of  the  pulley;  it 
will  cut  the  hyperbola  CBD'hom  the  cone  PEG,  which  forms  one  side  of 
the  groove.  The  cord  will  he  upon  the  pulley  from  B  to  7,  where  it  will 
leave  the  hyperbola  on  a  tangent.  If  the  tangent  at  I  falls  well  within 
the  edge  of  the  pulley  at  C,  the  groove  is  deep  enough.  It  will  usually 
be  sufficient  to  draw  a  straight  line,  as  ab  (Fig.  64),  and  see  that  it  falls 
well  inside  of  the  point  corresponding  to  C  in  Fig.  65. 


FIG.  65 

69.  Drum  or  Barrel.     When  a  cord  does  not  merely  pass  over  a 
pulley,  but  is  made  fast  to  it  at  one  end  and  wound  upon  it,  the 
pulley  usually  becomes  what  is  called  a  drum  or  barrel.     A  drum  for 
a  round  rope  is  cylindrical  and  the  rope  is  wound  upon  it  in  helical  coils. 
Each  layer  of  coils  increases  the  effective  radius  of  the  drum  by  an 
amount  equal  to  the  diameter  of  the  rope.     A  drum  for  a  flat  rope 
has  a  breadth  equal  to  that  of  the  rope,  which  is  wound  upon  itself  in 
single  coils,  each  of  which  increases  the  effective  radius  by  an  amount 
equal  to  the  thickness  of  the  rope. 

70.  Wire  Ropes.     Wire  rope  is  well  adapted  for  the  transmission 
of  large  powers  to  great  distances,  as  for  instance  in  cable  and  in- 
clined railways.     Its  rigidness,   great  weight,   and  rapid  destruction 
due  to  bending,  however,  unfit  it  for  use  in  mill  service,  where  the 
average  speed  of  rope  is  about  4000  ft.  per  minute.     As  the  easiest 
way  to  break  wire  is  by  bending  it,  ropes  made  of  it,  by  any  method 


BELTS,   ROPES  AND  CHAINS 


55 


whatsoever,  have  proved  unsatisfactory  for  drives  of  short  centers  and 
high  speed  unless  the  diameters  of  the  sheaves  are  large  enough  to  avoid 
bending  the  rope  to  strain  it  above  the  elastic  limit. 

Wire  ropes  will  not  support  without  injury  the  lateral  crushing  due 
to  the  V-shaped  grooves ;  hence  it  is  necessary  to  construct  the  pulleys 
with  grooves  so  wide  that  the  rope  rests  on  the  rounded  bottom  of  the 
groove,  as  shown  in  Fig.  66,  which  shows  a  section  of  the  rim  of  a  wire- 
rope  pulley.  The  friction  is  greatly  increased,  and  the  wear  of  the 


FIG.  66 

rope  diminished,  by  lining  the  bottom  of  the  groove  with  some  elastic 
material,  as  gutta-percha,  wood  or  leather  made  up  in  short  sections 
and  forced  into  the  bottom  of  the  groove. 

71.  Chains  are  frequently  used  as  connectors  between  parallel 
axes  and  also  for  conveying  and  hoisting  machinery  and  for  other 
similar  purposes.  The  wheels  over  which  chains  run  are  called  sprockets 
and  have  their  surfaces  shaped  to  conform  to  the  type  of  chain  used. 

Chains  may  be  classified  as  follows: 

1°  Hoisting  chains 

f  Detachable  or  Hook  Joint 
{  Closed  Joint 


2°  Conveyor  chains 


3°  Power  transmission  chains 


Block 
Roller 
Silent 


72.  Hoisting  Chains.     The  most  common  form  of  hoisting  chain 
consists  of  solid  oval  links  as  shown  in  Fig.  67.     The  form  of  sprocket 
used  for  such  a  chain  is  evident  from  the  figure. 

73.  Conveyor  Chains  may  be  of  the  detachable  or  hook  joint  type 
as  shown  in  Fig.  68,  or  of  the  closed  joint  type  illustrated  in  Fig.  69. 


56 


ELEMENTS  OF  MECHANISM 


FIG.  67 


I O 


f 


FIG.  68 


iii 


Squared 
./       X 


FIQ.  69 


FIG.  70 


BELTS,  ROPES  AND  CHAINS 


57 


The  design  of  the  sprocket  teeth  is  largely  empirical,  care  being 
taken  to  have  the  teeth  so  shaped  and  spaced  that  the  chain  will  run 
on  to  and  off  from  the  sprockets  smoothly  and  without  interference 
even  after  it  has  stretched  or  worn  somewhat.  Chains  of  this  general 
class  are  often  used  for  transmitting  power  at  low  speeds,  as  in  agricul- 
tural machinery.  They  are  usually  made  of  malleable  cast  links  and 


FIG.  71 

do  not  have  the  smooth  running  qualities  of  the  more  carefully  made 
chains. 

74.  Power  Transmission  Chains.     This  class  includes  the  three 
types  known  as  block,  roller  and  silent.     The  chains  are  made  of 
steel,  accurately  machined,  with  wearing  parts  hardened,  and  run  on 
carefully  designed  sprockets.     In  the  following  discussion  no  attempt 
is  made  to  give  an  exhaustive  treatment  of  the  subject,  but  merely 
to  give  some  idea  of  the  character  of  the  three  types  and  some  of  the 
points  which  need  to  be  considered  in  their  design. 

75.  Block  Chains.     Fig.   70  shows  a  block  chain  made  by  the 
Diamond  Chain  &  Mfg.  Co. 


58 


ELEMENTS  OF  MECHANISM 


FIG.  73 


Chains  of  the  block  type  are  less  expensive  to  make  than  the  roller 
or  silent  chains  and  are  used  for  the  transmission  of  power  at  com- 
paratively low  speeds.  They  are  also  used  to  some  extent  as  conveyor 
chains  and  for  other  purposes  in  place  of  the  malleable  chains  of  class  2. 


BELTS,  ROPES  AND  CHAINS 


59 


Fig.  71  shows  a  block  chain  in  place  on  the  driving  sprocket.  Atten- 
tion is  called  to  the  way  in  which  the  links  swing  into  position  as  they 
approach  the  sprocket  and  swing  out  as  they  leave.  A  method  of 
laying  out  the  sprocket  teeth  is  indicated  on  the  same  figure.  The 
proportions  for  the  teeth  here  shown  are  those  recommended  by  Mr. 
B.  D.  Pinkney  in  "Machinery,"  January,  1916. 

76.  Roller  Chains.  Fig.  72  illustrates  a  form  of  roller  chain  similar 
to  one  made  by  the  Diamond  Co.,  and  Fig.  73  shows  the  same  chain 


FIG.  74 

in  place  on  the  sprocket.  The  method  of  laying  out  the  sprocket  teeth 
in  Fig.  73  is  the  one  recommended  by  the  Diamond  Co.,  for  roller  chain 
sprockets. 

77.  Calculations  for  Chain  Length.  This  paragraph  and  the  two 
following  with  Figs.  74,  75  and  76  are  taken  directly  from  "Power 
Chains  and  Sprockets"  published  by  the  Diamond  Chain  &  Mfg.  Co. 

D  =  Distance  between  centers. 
A  =  Distance  between  limit  of  contact. 
R  =  Pitch  radius  of  large  sprocket. 
r  =  Pitch  radius  of  small  sprocket. 
N  =  Number  of  teeth  on  large  sprocket. 
n  =  Number  of  teeth  on  small  sprocket. 
P  =  Pitch  of  chain  and  sprocket. 
(180°  +  2  a)  =  Angle  of  contact  on  large  sprocket. 
(180°  —  2  a)  =  Angle  of  contact  on  small  sprocket. 


a  =  sin 


A  =  D  cos  a. 


60 


ELEMENTS  OF  MECHANISM 


Total  length  of  chain. 


360 


NP 


180  -  2  a 
360 


cos  a. 


(32) 


78.   Calculations  for  Diameters  of  Sprockets  for  Block  Chains. 

N  =  Number  of  teeth. 

b  =  Diameter  of  round  part  of  chain  block  (usually  0.325). 
B  =  Center  to  center  of  holes  in  chain  block  (usually  0.4). 
A  =  Center  to  center  of  holes  in  side  links  (usually  0.6). 

180° 


a  =• 


N 


sin  a 


B  . 

-7  +  cos  a 

A 


Pitch  diameter  = 


sin  B 

Outside  diameter  =  Pitch  diameter  +  b. 
Bottom  diameter  =  Pitch  diameter  —  6. 


(33) 

(34) 

(35) 


In  calculating  the  diameter  of  sprocket  wheels,  the  bottom  diameter 
is  the  most  important. 


FIG.  75 


FIG.  76 


79.  Calculations  for  Diameters  of  Sprockets  for  Roller  Chains. 
Referring  to  Fig.  76, 


BELTS,  ROPES  AND  CHAINS 

N  =  Number  of  teeth  in  sprocket. 
P  =  Pitch  of  chain. 
D  =  Diameter  of  roller. 
180° 
N 
P 


61 


a  = 


Pitch  diameter  = 


(36) 


sin  a 

Outside  diameter  =  Pitch  +  D.  (37) 

Bottom  diameter  =  Pitch  -  D.  (38) 

80.  Silent  Chains.     None  of  the  above  mentioned  chains  can  be 
run  at  high  speed  without  noise.     There  are  now  in  use  several  makes 
of  chains  known  as  Silent  Chains  which  run  satisfactorily  at  high 
speeds  and  which  adapt  themselves  to  the  sprocket  after  the  pitch 
of  the  chain  has  increased  due  to  wear.     Two  examples  will  serve  to 
illustrate  this  type  of  chain. 

81.  Renold  Silent  Chain.     Fig.  77  shows  a  chain  developed  by 
Hans  Renold.     It  consists  of  links  C  of  a  peculiar  form  with  straight 


FIG.  77 

bearing  edges  a,  6,  which  run  over  cut  sprocket-wheels  with  straight- 
sided  teeth  whose  angles  vary  with  the  diameter  of  the  wheel.  The 
chain  may  be  made  any  convenient  width,  the  pins  binding  the  whole 
together.  One  sprocket  of  each  pair  is  supplied  with  flanges  to  retain 
the  chain  in  place.  The  upper  drawing  shows  a  new  chain  in  position 
on  its  sprocket,  the  bearing  parts  of  the  links  being  on  the  straight 
edges  of  the  links  only,  not  on  the  tops  or  roots  of  the  teeth.  The 
chain  thus  adjusts  itself  to  the  sprocket  at  a  diameter  corresponding 
with  its  pitch,  and  as  any  tooth  comes  into  or  out  of  gear  there  is  neither 


62  ELEMENTS  OF  MECHANISM 

slipping  nor  noise.    The  lower  figure  shows  the  position  taken  by  a 
worn  chain  of  increased  pitch  on  the  same  wheel. 

82.  Morse  Rocker- Joint  Chain.  This  chain  (Fig.  78)  eliminates 
the  sliding  friction  of  the  rivets  as  the  chain  bends  around  the  sprocket. 
Instead  of  the  ordinary  pin  bearing  a  rocking  bearing  is  provided  at 
each  joint.  The  following  description,  with  slight  changes,  is  taken 
from  the  catalogue  of  the  Morse  Chain  Co.  Two  pins  are  employed 
at  each  joint;  the  left  hand  pin  a  is  called  the  seat  pin  and  the  right 
hand  pin  b  the  rocker.  Each  is  securely  held  in  its  respective  end  of 
the  link.  The  seat  pin  has  a  plane  surface  against  which  the  edge  of 
the  rocker  pin  rocks  or  rolls  when  the  chain  goes  on  and  off  the  sprockets. 
The  joint  is  so  designed  that  the  pressure  due  to  tension  of  driving  will 
be  taken  on  a  flat  surface  when  in  between  the  sprockets. 


FIG.  78 

Fig.  78  shows  the  chain  on  a  driving  sprocket  running  in  the  direc- 
tion indicated  by  the  arrows.  The  angle  of  the  tooth  to  the  line  of 
the  pull  and  any  centrifugal  force  that  may  exist  both  tend  to  keep 
the  link  out  to  its  true  pitch  diameter  during  the  revolution  of  the 
wheel;  it  will  fall  below  this  point  only  when  the  pull  of  the  slack  side 
of  the  chain  is  greater  than  the  forces  in  the  opposite  direction. 

From  this  it  will  be  seen  that  there  are  two  forces  definitely  opera- 
tive to  keep  the  chain  in  its  proper  pitch  contact  with  the  wheels  by 
causing  it  to  assume  a  larger  and  larger  circle  as  the  chain  lengthens  in 
pitch;  thus,  the  driving  load  continues  to  be  distributed  over  a  large 
number  of  teeth. 

The  climbing,  which  compensates  for  the  increase  of  pitch,  is  gradual, 
easily  noticed  in  the  running  drive,  does  not  decrease  the  efficiency 
of  the  transmission,  and,  as  the  chain  lengthens  and  approaches  the 
top  of  the  teeth,  gives  fair  warning  of  the  necessity  of  replacement  or 
repairs  of  the  chain. 


CHAPTER  IV 

TRANSMISSION  OF  MOTION  BY  BODIES  IN  PURE 
ROLLING  CONTACT 

83.  Pure  Rolling  Contact  consists  of  such  a  relative  motion  of  two 
lines  or  surfaces  that  the  consecutive  points  or  elements  of  one  come  suc- 
cessively into  contact  with  those  of  the  other  in  their  order.     There  is 
no  slipping  between  two  surfaces  which  have  pure  rolling  contact,  that 
is,  all  points  in  contact  have  the  same  linear  speed. 

Two  bodies  may  be  rotating  on  their  respective  axes,  so  arranged 
that,  by  pure  rolling  contact,  one  may  cause  the  other  to  turn  with  an 
angular  speed  bearing  a  definite  ratio  to  the  angular  speed  of  the  driver. 
This  speed  ratio  may  be  constant  or  variable,  depending  upon  the  forms 
of  the  two  bodies.  The  axes  may  be  parallel,  intersecting,  or  neither 
parallel  nor  intersecting  *  The  present  chapter  will  consider  the  cases  of 
parallel  axes  connected  by  cylinders  giving  constant  speed  ratio,  inter- 
secting axes  connected  by  cones  giving  constant  speed  ratio  and  parallel 
axes  connected  by  bodies  of  irregular  outline,  giving  variable  speed  ratio. 

The  connection  between  non-parallel,  non-intersecting  axes  will  be 
discussed  in  connection  with  the  subject  of  gearing. 

84.  Cylinders  Rolling  Together  without  Slipping.     External  Con- 
tact.    In  Fig.  79  let  A  be  a  cylinder  fast  to  the  shaft  S  and  B  a  cylinder 
fast  to  the  shaft  Si.     Assume  that  the  shafts  are  held  by  the  frame  so 
that  their  centers  are  at  a  distance  apart  just  equal  to  the  sum  of  the  radii 
of  the  two  cylinders;  that  is,  R  +  Ri  =  C.     Then  the  surfaces  will  touch 
at  P.     Suppose  also  that  the  nature  of  the  surfaces  of  the  cylinders  is 
such  that,  as  they  turn  on  their  respective  axes,  there  can  be  no  slipping 
of  one  surface  on  the  other.     Then  the  surface  speed  of  A  must  be  equal 
to  that  of  B,  and  A  and  B  must  turn  in  such  directions  relative  to  each 
other  that  the  element  on  A  which  is  in  contact  with  B  is  moving  in  the 
same  direction  as  the  element  on  B  which  it  touches.     (Notice  the 
arrows  in  the  figure,  the  full  arrows  belonging  together  and  the  dotted 
arrows  together.) 

If  A  makes  N  r.p.m.  and  B  makes  Ni  r.p.m., 

Surface  speed  of  A  =  2  irRN 
and  Surface  speed  of  B  =  2  irRiNi. 

*  In  the  case  of  axes  which  are  neither  parallel  nor  intersecting  the  coinciding 
elements  of  the  rolling  bodies  may  slide  on  each  other  in  the  direction  of  their  length, 
so  that  the  contact  is  not  pure  rolling  in  a  strict  sense. 

63 


64  ELEMENTS  OF  MECHANISM 

Therefore,  if  the  surface  speed  of  A  equals  the  surface  speed  of  B, 


\    or    ^-=22.  (39) 

Or,  in  other  words,  the  angular  speeds  of  two  cylinders  which  roll  together 
without  slipping  are  inversely  proportional  to  the  radii  of  the  cylinders. 
It  will  be  noticed  that  this  principle  is  the  same  as  that  shown  in  the 

preceding  chapter  applied  to 
cylinders  connected  by  a  belt 
or  other  flexible  connector. 

85.  Solution  of  Problems 
on  Cylinders  in  External 
Contact.  In  Fig.  79  suppose 
C,  N  and  NI  are  known;  re- 
quired to  find  the  diameters 


of  the  two  cylinders. 
Eq.  (39), 


From 


R 


R  = 


RiN, 


N-R.P.M 


FIG.  79 


Ri      N  N 

It  is  known  also  that  R  +  RI 
=  C.   R  and  H\  can,  therefore, 
be  found  by  solving  these  as 
simultaneous  equations. 
The  same  result  may  also  be  found  by  a  simple  graphical  construction 

as  follows:  Draw  the  line  SSi,  Fig.  80,  making  its  length  equal  to  the 

distance  between  the  centers  of  the  shafts  (corresponding  to  C,  Fig.  79). 

This  would,  in  most  cases,  have  to 

be  drawn    at    some    reduced    scale. 

From   S   draw   a   line   SV,    making 

any  angle  with  SSi.     From  S  lay  off 

the  distances  SK  equal  to  NI  linear 

units  and  KT  equal  to  N  linear  units. 

The   line    ST  is   then  divided   into 

two  parts   SK   and  KT  such   that 

jr~,  =  -Tr1'    Now  connect  T  with  Si  and  from  K  draw  a  line  parallel  to 
TSi,  cutting  SSi  at  P.     Then,  from  the  similar  triangles  SKP  and 

orr>          ar?-  AT 

STSi,  T^T  =  IFTF  =  ifr1*    Therefore,  SP  will  be  the  radius  R  and  PSi  the 

.roi       /Y  L          N 

radius  R\. 

Example  18.  Two  shafts  A  and  B  are  16  ins.  on  centers.  A  is  to  turn  50  times  in 
a  minute  and  B  150  times  in  a  minute.  What  must  be  the  size  of  the  cylinders  to 
connect  them  if  they  are  to  turn  in  opposite  directions? 


FIG.  80 


TRANSMISSION  OF  MOTION 


65 


A 

50  R.P.M 


150  R. P.M. 


Calculation.     From  Eq.  (39), 

Radius  of  A  _  Turns  of  B  per  minute  _  150  _  3^ 
Radius  of  B  ~  Turns  of  A  per  minute  ~  50   ~  1 ' 
or, 

Radius  of  A  =  3  X  radius  of  B. 
Also 

Radius  of  A  +  radius  of  B  =  16  ins. 
Therefore, 

Radius  of  B  =  4  ins. 
and 

Radius  of  A  =  3  X  4  =  12  ins. 

Graphical  Solution.  In  Fig.  81 
draw  the  line  AB  equal  to  16  ins. 
at  some  reduced  scale. 

From  A  draw  the  line  AV  at  FIG.  81 

any  angle.     Lay  off  AK  equal  to 

150  units.  Lay  off  KT  =  50  of  the  same  units.  Join  T  with  B  and  draw  KP  parallel 
to  TB.  Then  BP  will  be  found  to  measure  4  ins.  and  AP  12  ins.,  making  proper 
allowance  for  the  scale  at  which  AB  was  drawn. 

Example  19.  A  cylinder  18  ins.  diameter  on  a  shaft  A  making  75  r.p.m.  drives  by 
rolling  contact  a  cylinder  on  another  shaft  B,  the  second  cylinder  being  4|  ins.  diam- 
eter. How  fast  does  B  turn  if  the  shafts  turn  in  opposite  directions? 

r.p.m.  of  B  _  Diam.  of  cylinder  on  A  _  18  _  4 
r.p.m.  of  A  ~  Diam.  of  cylinder  on  B      4|       1 
or  r.p.m.  of  B  =  4  X  r.p.m.  of  A  =  4  X  75  =  300. 

Graphical  Solution.     (Fig.  82.)     Draw  the*  line  AB  equal  in  length  to  the  sum  of 


Calculation. 


18      4- 
the  radii  of  the  two  cylinders  =  -5-  +  -^ 


ins. 


On  AB  locate  the  point  P  9  ins.  from  A  (therefore  2£  his.  from  B).  From  B,  the 
center  of  the  cylinder  whose  speed  is  to  be  found,  draw  a  line  BV  at  any  angle  and 

lay  off  on  this  line  BK  equal  to 

.fffftt  \  .,  I  _\xv  75  units  (that  is,  speed  of  A). 
Join  K  with  P  and  through  A 
draw  a  line  parallel  to  PK  cutting 
BV  at  T.  Then  the  units  in  KT 
will  show  the  speed  of  B. 


*  n 
75B.P.M.W 


FIG.  82 


86.  Cylinders  Rolling  To- 
gether without  Slipping. 
Internal  Contact.  In  Fig. 
83,  where  the  lettering  corre- 
sponds to  that  of  Fig.  79,  the 
cylinder  A  is  hollow  with  B  inside  it,  so  that  the  contact  is  between  the 
inner  surface  of  A  and  the  outer  surface  of  'B.  This  is  called  internal 
contact.  The  same  mathematical  reasoning  will  apply  here  as  in  Fig. 
79,  and  Eq.  (39)  will  hold  true.  The  distance  between  centers  now, 


66 


ELEMENTS  OF  MECHANISM 


however,  is  equal  to  R  —  Ri  instead  of  R  +  RI.     The  two  cylinders  in 

Fig.  83  will  turn  in  the  same  direction  instead  of  in  opposite  directions. 

87.   Solution  of  Problems  on  Cylinders  in  Internal  Contact.     In 

Fig.  83  if  C,  N  and  Ni  are  known,  to  find  the  diameters  of  the  cylinders. 
From  Eq.  (39), 


It  is  also  known  that  R  —  Ri  =  C.    These  may  be  solved  as  simul- 
taneous equations  to  find  R  and  R\9  and  therefore  the  diameters. 

The  graphical  solution  of  prob- 
lems on  cylinders  rolling  in 
internal  contact  is  similar  in 
principle  to  that  shown  in  Fig.  80 
for  external  contact.  Fig.  84 
shows  the  construction  for  in- 
ternal contact. 


N-R.P.M. 


FIG.  84 
j -R.P.M.. 

Example  20.   Two  shafts  A  and  B 
are  8  ins.  on  centers  and  are  to  be 

connected  by  rolling  cylinders  to  turn  in  the  same  direction,  A  to  make  20  r.p.m. 
and  B  to  make  60  r.p.m.     Find  the  diameters  of  the  cylinders. 
Calculation.     From  Eq.  (39), 

Rad.^1  =  r.p.m.  B  =  60  =  3 
Had.  B  ~  r.p.m.  A      20      1 
Had.  A  =  3  X  rad.  B, 
Rad.  A  -  rad.  B  =  8  ins. 
3  rad.  B  -  rad.  B  =  8  ins. 
2  rad.  B  =  8  ins. 
Rad.  B  =  4  ins. 

Rad.  A  =  3  X  rad.  5  =  3X4  ins.  =  12  ins. 
Graphical  Solution.     Fig.  85  shows  the  graphical  solution  for  this  problem. 

Example  21.   A  cylinder  24  in.  diameter  on  a  shaft  A  making  60  r.p.m.  drives 
by  rolling  contact  a  cylinder  on  another  shaft  B,  the  second  cylinder  being  6  in. 


or 
also 

or 
and 


TRANSMISSION  OF  MOTION 


67 


diameter.     The  shafts  turn  in  the  same  direction.     How  fast  does  B  turn  and  how 
far  apart  are  the  shafts? 

Rev.  B      Diam.  cyl.  on  A      24      4 
Calculation.  -^-^  =  Diam.  cyL  on  B  =  ~Q   =  f ' 

R.p.m.  of  B  =  4  X  r.p.m.  of  A  =  4  X  60  =  240  r.p.m. 
and  Dist.  between  centers  =  rad.  A  —  rad.  B 

or  Dist.  between  centers  =  12  —  3  =  9  ins. 

Graphical  Solution.    Fig.  86  shows  the  graphical  solution  of  Example  21. 


20  P.P. 


MS.?.*.  WftP.J<l 


FIG.  85 


FIG.  86 


88.   Cones  Rolling  Together  without  Slipping.     External  Contact. 

In  the  preceding  discussion  relating  to  cylinders,  the  shafts  were  neces- 
sarily parallel.  It  is  often  necessary  to  connect  two  shafts  which  lie  in 
the  same  plane  but  make  some  angle  with  each  other.  This  is  done  by 
means  of  right  cones  as  shown  in  Fig.  87  or  frusta  of  cones  as  shown  in 
Fig.  88,  the  cones  having  a  common  apex.  The  same  reasoning  applies 
to  the  ratio  of  speeds  at  the  base  of  the  cones  as  to  the  circles  repre- 
senting the  cylinders  in  Fig.  79.  That  is, 

£,=§•  <«> 


But 

Therefore, 


Ri  =  OP  sin  POCi    and    R  =  OP  sin  POC. 
Ri  =  OP  sin  POCi  =  sin  POCi 
R  ~  OP  sin  POC  ~  sin  POC  ' 


Substituting  this  expression  in  Eq.  (40), 

N_      sin  POCi 
Ni  ~  sin  POC  ' 

Therefore,  the  angular  speeds  of  two  cones  rolling  together  without 
slipping  are  inversely  as  the  sines  of  the  half  angles  of  the  cones. 

89.  Solution  of  Problems  on  Cones  in  External  Contact.  The  law 
stated  in  the  previous  paragraph  may  be  made  use  of  to  calculate  the 
vertical  angles  of  the  cones  when  the  angle  between  the  axes  and  the 
speed  ratio  are  known, 


68 


ELEMENTS  OF  MECHANISM 


Referring  to  Fig.  88,  let  angle  COCi  =  0, 

POC  =  a    and    angle  POCi  =  ft. 
N       sin  ft  sin  ft  sin  ft 


Then          £• 


sin  a      sin  (0  —  ft)      sin  0  cos  ft  —  cos  0  sin  ft 

Sin  0 

Tan/3 


Cos  ft 
Sin  6  —  cos  0 


Whence 


sin  ft     Sin  0  —  cos  0  tan 
cos  0 

Tan/3  = 


Sin0 


In  similar  manner 


Tan  a  = 


+  cos* 
Sin0 


(42) 


N 


+  COS0 


FIG.  87 


Graphical  Construction.  In  Fig.  89,  /S  and  Si  are  two  shafts  which  are 
to  be  connected  by  rolling  cones  to  turn  as  indicated  by  the  arrows. 
Their  center  lines  meet  at  0.  S  is  to  make  N  r.p.m.  and  Si  is  to  make 
Ni  r.p.m.  Required  to  find  the  line  of  contact  of  two  cones  which  will 
connect  the  shafts,  and  to  draw  a  pair  of  cones. 


TRANSMISSION  OF  MOTION 
0 


69 


X% 

/     I    '  \  \ 
/  \    \ 

M-R.P.M.        /          A*-t-A      XN 


N,-R.P.M. 


FIG.  88 


FIG.  89 


70 


ELEMENTS  OF  MECHANISM 


Draw  a  line  parallel  to  OA,  on  the  side  toward  which  its  direction 
arrow  points,  at  a  distance  from  OA  equal  to  Ni  units.  Draw  a  similar 
line  parallel  to  OB,  N  units  distant  from  OB.  These  two  lines  intersect 
at  K.  A  line  drawn  through  0  and  K  will  be  the  line  of  contact  of  the 
required  cones.  Select  any  point  P  on  OK  and  from  P  draw  lines  per- 
pendicular to  AO  and  BO  meeting  AO  and  BO  at  M  and  M\t  respectively. 
Produce  these  lines,  making  ME  =  MP  and  MiJ  =  MiP.  Draw  HO 
and  JO.  Then  OPH  and  OPJ  are  cones  of  the  proper  relative  sizes  to 
connect  S  and  Si  to  give  the  required  speeds. 

If  the  point  P  had  been  chosen  nearer  to  0,  the  cones  would  have  had 
smaller  diameters  at  their  bases  but  the  ratio  of  the  diameters  would 
have  been  the  same,  or,  if  P  had  been  chosen  farther  away  from  0,  the 
bases  would  have  been  larger  but  still  of  the  same  ratio.  If  frusta  of 
cones  are  desired,  the  cones  can  be  cut  off  anywhere,  as  shown  by  the 
dotted  lines  FE  and  FG. 

Example  22.  Two  shafts  S  and  Si,  Fig.  90,  in  the  same  plane,  make  an  angle  of 
105°  with  each  other.  S  is  to  turn  90  times  per  minute  and  Si  30  times  per  minute. 
A  cone  on  A  having  a  base  f  ins.  diameter  is  to  roll  with  a  cone  on  B  to  give  the  re- 
quired speeds;  directions  of  rotation  are  to  be  as  shown.  To  find  the  diameter  of 
the  base  of  the  cone  on  B  and  to  draw  the  cones. 


MR.  P.M.' 


FIG.  90 

Solution.  Draw  a  line  parallel  to  S  30  units  distant  from  S,  and  a  line  parallel 
to  Si  90  units  distant  from  Si.  These  lines  intersect  at  K',  then  KO  is  the  element  of 
contact.  Since  the  base  of  the  smaller  cone  is  to  be  |  ins.  diameter,  find  a  point  P  on 
OK  which  is  |  in.  from  S.  Through  this  point  draw  the  bases  of  the  cones  perpendic- 
ular to  S  and  Si  in  the  same  way  as  explained  for  Fig.  89. 

Calculation  of  Angles  a  and  /3. 
Using  Equation  (42), 

Tan/3  = 


£+003, 


where 


Tan/3  = 


TRANSMISSION  OF  MOTION 

=  30,      N  =  90, 
=  12.9530. 


71 


6  =  105°, 

0.9659 


+  0.2588 

0  =  85°  25'  nearly. 
2  £  =  170°  50'  nearly  =  angle  at  apex  of  cone  on  Si. 

a.  =  105°  -  85°  25'  =  19°  35'. 
2  a  =  39°  10'  =  angle  at  apex  of  cone  on  S. 


FIG.  91 


90.   Cones  Rolling  Together  without  Slipping.     Internal  Contact. 

The  combination  of  speed  ratio,  directional  relation,  and  angle  between 
axes  may  be  such  as  to  require  the  cones  to  be  arranged  for  internal 


72 


ELEMENTS  OF  MECHANISM 


contact.  This  construction  is  not  very  often  met  in  practice,  but  the 
necessity  for  it  sometimes  occurs  and  it  is  well  to  be  familiar  with  the 
problem.  Fig.  91  shows  the  principle  involved.  Corresponding  points 
are  lettered  the  same  as  in  Fig.  87,  and  the  same  equations  apply  for 
finding  the  ratio  of  speeds  of  the  two  shafts. 

It  is  not  always  possible  to  predict  from  given  conditions  whether  the 
contact  will  be  external  or  internal. 

91.  Solution  of  Problems  on  Cones  Rolling  in  Internal  Contact. 
The  same  general  methods  apply  to  solution  of  problems  on  cones  in 
internal  contact  as  in  the  case  of  external  contact.  It  should  be  noted, 
however,  that  since  6  =  a  —  0  the  equation  corresponding  to  Equation 
(42)  becomes  _Sin*_  .  (43) 


Example  23.  The  axes  of  two  shafts  S  and  Si  intersect  at  an  angle  of  45°.  S 
makes  15  r.p.m.  and  Si  makes  60  r.p.m.  The  shafts  are  to  be  connected  by  two  cones 
rolling  together  without  slipping.  The  connection  to  be  such  that  each  turns  as 
shown  by  the  arrows.  To  draw  the  outline  of  the  cones  and  calculate  the  angles 
at  the  apices. 


FIG.  92 

Graphical  Solution.  (Fig.  92.)  Draw  the  center  lines  of  the  shafts,  SO  and  SiO 
intersecting  at  45°  and  put  on  the  direction  arrows  as  shown.  Draw  line  MR 
parallel  to  SiO  and  15  units  distant  from  it  on  the  side  toward  which  the  direction 


TRANSMISSION  OF   MOTION 


73 


arrow  points.  In  a  similar  way  draw  TK  parallel  to  SO  and  60  units  distant  from  it. 
The  point  C  where  MR  and  TK  intersect  will  be  the  point  through  which  the  element 
of  contact  CO  is  drawn.  Having  the  element  of  contact  the  cones  may  be  drawn  as 
indicated  in  previous  examples  and  are  found  to  be  in  internal  contact. 


Calculations  of  Angles  of  Apices. 
Using  Equation  (43) 


Tan  ft  = 


Sin 


F 


—  COS0 


Where 


a  =  45  —  12°  7 


=  60,          N»  15, 


6  =  45°, 
Tan  ft  = 

ft  =  12°  7'  nearly.          2  ft  =  24°  14'  =  angle  at  apex  of  cone  on  Si. 

32°  53'  nearly.    2  a  =  65°  46'  =  angle  at  apex  of  cone  on  S. 


0.7071 
M  -  0.7071       U'J14'- 


92.  Rolling  Cylinder  and  Sphere.  —  Fig.  93  shows  an  example  of  a 
rolling  cylinder  and  sphere  as  used  in  the  Coradi  planimeter.  The  seg- 
ment of  the  sphere  A  turns  on  an  axis  ac  passing  through  a,  the  center 


FIG.  93 

of  the  sphere.  The  cylinder  B,  whose  axis  is  located  in  a  plane  also 
passing  through  the  center  of  the  sphere,  is  supported  by  a  frame  pivoted 
at  e  and  is  held  to  the  cylinder  by  a  spring,  not  shown.  The  frame 
pivots  e  are  movable  about  an  axis  at  right  angles  to  ac  and  passing 
through  a,  the  center  of  the  sphere.  When  the  roller  is  in  the  position 
B  with  its  axis  at  right  angles  to  ac,  the  turning  of  the  sphere  produces 
no  motion  of  B'}  when,  however,  the  roller  is  swung  so  that  its  axis 
makes  an  angle  bac\  with  its  former  position,  as  shown  at  BI  by  dotted 
lines,  the  point  of  contact  is  transferred  to  ci  in  the  perpendicular  from 
a  to  the  roller  axis.  If  now  we  assume  the  radius  of  the  roller  =  R,  the 
relative  motion  of  roller  and  sphere,  in  contact  at  Ci,  is  the  same  as  that 


74 


ELEMENTS  OF   MECHANISM 


of  two  circles  of  radii  R  and  bci  respectively.  Transferring  the  point  of 
contact  to  the  opposite  side  of  ab  will  result  in  changing  the  directional 
relation  of  the  motion.  The  action  of  this  device  is  purely  rolling  and 
but  very  little  force  can  be  transmitted.  It  is  used  only  in  very  delicate 
mechanisms. 

93.  Disk  and  Roller.  —  If  in  Fig.  93  the  radius  of  the  sphere  ac  is 
assumed  to  become  infinite  and  the  roller  B  to  be  replaced  by  a  sphere 
of  the  same  diameter  turning  on  its  axis,  the  result  will  be  a  disk  and  roller 
as  shown  in  Fig.  94,  where  AA  represents  the  disk  and  B  the  roller,  made 
up  of  the  central  portion  of  the  sphere. 

If  we  suppose  the  rotation  of  the  disk  to  be  uniform,  the  velocity 
ratio  between  B  and  A  will  constantly  decrease  as  the  roller  B  is  shifted 

nearer  the  axis  of  A,  and 
conversely.  If  the  roller  is 
carried  to  the  other  side  of 
the  axis,  it  will  rotate  in  the 
opposite  direction  to  the  first. 
This  combination  is  some- 
times used  in  feed  mechan- 
isms for  machine  tools,  where 
it  enables  the  feed  to  be 
adjusted  and  also  reversed  by 
simply  adjusting  the  roller 
on  the  shaft  CC.  If  possible 
the  roller  should  drive. 

94.  Friction  Gearing.     Rolling  cylinders  and  cones  are  frequently 
used  to  transmit  force,  and  constitute  what  is  known  as  friction  gearing. 
In  such  cases  the  axes  are  arranged  so  that  they  can  be  pressed  together 
with  considerable  force,  and,  in  order  to  prevent  slipping,  the  surfaces 
of  contact  are  made  of  slightly  yielding  materials,  such  as  wood,  leather, 
rubber  or  paper,  which,  by  their  yielding,  transform  the  line  of  contact 
into  a  surface  of  contact  and  also  compensate  for  any  slight  irregulari- 
ties in  the  rolling  surfaces.     Frequently  only  one  surface  is  made  yield- 
ing, the  other  being  usually  made  of  iron.     As  slipping  is  likely  to  take 
place  in  these  combinations,  the  velocity  ratio  cannot  be  depended  upon 
as  absolute. 

When  rolling  cylinders  or  cones  are  used  to  change  sliding  to  rolling 
friction,  that  is,  to  reduce  friction,  their  surfaces  should  be  made  as  hard 
and  smooth  as  possible.  This  is  the  case  in  roller  bearings  and  in  the 
various  forms  of  ball  bearings  where  spheres  are  arranged  to  roll  in 
suitably  constructed  races,  all  bearing  surfaces  being  made  of  hardened 
steel  and  ground. 


FIG.  94 


TRANSMISSION  OF  MOTION 


75 


FIG.  95 


Friction  gearing  is   utilized   in   several   forms   of  speed-controlling 

devices,  among  which  the  following  are  good  examples : 

Fig.  95  shows  the  mechanism  of  the  Evans  friction  cones,  consisting 

of  two  equal  cones  A  and  B  turning  on  parallel  axes  with  an  endless 

movable  leather  belt  C  in  the  form  of  a  ring  running  between  them, 

the  axis  of  B  being  urged 

toward  A   by  means  of 

springs  or  otherwise.    By 

adjusting  the  belt  along 

the  cones,  their  angular 

speed  ratio  may  be  varied 

at    will.     It    should    be 

observed  that  there  must 

be  some   slipping   since 

the  angular  speed  ratio 

varies  from  edge  to  edge 

of  the  belt,  the  resulting 

ratio  approaching  that  of  the  mean  line  of  the  belt.     A  leather-faced 

roller  might  be  substituted  for  the  belt  and  a  similar  series  of  speeds 

obtained,  the  cones  then  turning  in  the 
same  instead  of  in  opposite  directions. 

Fig.  96  shows,  in  principle,  another 
form,  made  by  the  Power  and  Speed 
Controller  Co.  Here  two  equal  rollers, 
C  and  D,  faced  with  a  yielding  material, 
are  arranged  to  run  between  two  equal 
hollow  disks  A  and  B.  The  rollers 
with  their  supporting  yokes  (only  one 
of  which  is  shown  in  the  elevation)  are 
arranged  as  indicated  in  the  figure  and 
are  made  by  a  geared  connection,  not 
shown,  to  turn  opposite  each  other  on 
the  vertical  yoke  axes,  s.  The  contour 
of  the  hollow  in  the  disks  must  thus  be 
an  arc  of  a  circle  of  radius  equal  that 
of  the  roller  drawn  from  s  as  a  center. 
If  now  the  disk  B  is  made  fast  to  the 
shaft,  and  A,  running  loose,  is  urged 
against  B  by  a  spring  or  otherwise,  a 

uniform  motion  of  A  may  be  made  to  give  varying  speeds  to  B  by 

turning  the  rollers  as  shown.     To  increase  the  power  two  sets  of  disks 

are  often  used. 


B 


FIG.  96 


76 


ELEMENTS  OF  MECHANISM 


Fig.  97  shows  the  Sellers  feed  disks  used  to  give  a  varying  angular 
speed  ratio  between  two  parallel  shafts,  one  of  them  controlling  the  feed 
on  a  machine. 

The  two  outer  wheels  are  thickened  on  their  peripheries  and  run 
between  two  convex  disks  BB  which  are  constantly  urged  together  by 

hidden  coil  springs  bearing  against  the 
spherical  washers  clearly  shown.  The 
disks  BB  are  supported  by  the  pivoted 
forked  arm  D.  If  now  the  disk  A  be 
given  a  uniform  angular  speed,  the 
disk  C  may  be  made  to  have  a  greater 
or  less  angular  speed  as  the  axis  of  the 
disks  BB  is  made  to  approach  or  recede 
from  A. 

In  Fig.  98  a  modified  form  of  the 
Sellers  disks  is  shown.  The  shaft  A 
is  driven  by  the  pulley  P  and  is  carried 
by  a  forked  arm  supplied  with  two 
bearings  CC  and  swinging  about  a 
point  near  the  center  of  the  pulley 
driving  P  by  means  of  a  belt.  The 
externally  rubbing  disks  B  are  free  to  slide  axially  on  the  shaft  A,  but 
turn  with  it  and  are  constantly  urged  apart  by  springs  clearly  shown. 
The  internally  rubbing  convex  disks  are  made  fast  to  the  driven  shaft 
by  set  screws.  To  vary  the  speed  of  D,  that  of  A  being  constant,  it 
is  only  necessary  to  vary  the  distance  between  the  shafts.  In  the 
position  shown  D  has  its  highest 
speed,  the  disks  rubbing  at  a. 
When  the  shaft  A  is  urged  in  the 
direction  of  the  arrows  the  rub- 
bing radius  on  B  is  diminished 
and  that  on  E  increased,  the 
disks  BB  approaching  each  other. 
The  disks  BB  may  be  made  solid 
and  one  of  the  disks  E  be  urged 
toward  the  other  by  a  spring  on 
its  hub,  which  would  simplify  the 
construction. 

95.  Grooved  Friction  Gearing. 
—  Another  form  of  friction  gearing  is  shown  in  Fig.  99.     Here  increased 
friction  is  obtained  between  the  rolling  bodies  by  supplying  their  sur- 
faces of  contact  with  a  series  of  interlocking  angular   grooves;   the 


FIG.  98 


TRANSMISSION  OF  MOTION 


77 


sharper  the  angle  of  the  grooves,  the  greater  the  friction  for  a  given 
pressure  perpendicular  to  the  axes;  both  wheels  are  usually  made  of 
cast  iron.  Here  the  action  is  no  longer  that  of  rolling  bodies;  but 
considerable  sliding  takes  place,  which  varies  with  the  shape  and  depth 
of  the  groove.  This  form  of  gearing  is  very  generally  used  in  hoisting 
machinery  for  mines  and  also  for  driving  rotary  pumps;  in  both 
cases  a  slight  slipping  would  be  an  advantage,  as  shocks  are  quite 
frequent  in  starting  suddenly  and  their  effect  is  less  disastrous  when 
slipping  can  occur. 

The  speed  ratio  is  not  absolute  but  is  substantially  the  same  as  that 
of  two  cylinders  in  rolling  contact  on  a  line  drawn  midway  between  the 


FIG.  99 


FIG.  100 


tops  of  the  projections  on  each  wheel,  they  being  supposed  to  be  in 
working  contact. 

96.  Rolling  of  Non-cylindrical  Surfaces. —  If  the  angular  speed  ratio 
of  two  rolling  bodies  is  not  a  constant,  the  outlines  will  not  be  circular. 
Whatever  forms  of  curves  the  outlines  take,  the  conditions  of  pure  roll- 
ing contact  should  be  fulfilled,  namely,  the  point  of  contact  must  be  on 
the  line  of  centers,  and  the  rolling  arcs  must  be  of  equal  length.  For 
example,  in  the  rolling  bodies  represented  by  Fig.  100  with  Oi  and 
o2  the  axes  of  rotation,  we  must  find  the  sum  of  the  radiants  in 
contact,  QIC  +  o2c,  equal  to  the  sum  of  any  other  pair,  as  Old  +  o2e,  Oif 
+  ozg;  and  also  the  lengths  of  the  rolling  arcs  must  be  equal,  cd  =  ce,  df 
=  eg.  This  will  cause  the  successive  points  on  the  curves  to  meet  on 
the  line  of  centers,  and  the  rolling  arcs,  being  of  equal  length,  will  roll 
without  slipping. 


78 


ELEMENTS  OF  MECHANISM 


There  are  four  simple  cases  of  curves  which  may  be  arranged  to  fulfill 
these  conditions: 

A  pair  of  logarithmic  spirals  of  the  same  obliquity. 
A  pair  of  equal  ellipses. 
A  pair  of  equal  hyperbolas. 
A  pair  of  equal  parabolas. 

We  shall  also  find  that  any  of  the  above  curves  may  be  transformed 
in  one  way  or  another  and  still  fulfill  the  conditions  of  perfect  rolling 

contact  while  allowing  a  wide 
range  of  variation  in  the  angu- 
lar speed  ratio. 

97.  The  Rolling  of  two 
Logarithmic  Spirals  of  Equal 
Obliquity.  —  Fig.  101  shows 
the  development  of  a  pair  of 
such  spirals,  where,  if  they  roll 
on  the  common  tangent  de, 
the  axes  Oi  and  o2  will  move 
along  the  lines  fg  and  hk 
respectively.  The  arcs  axc, 
cbi,  etc.,  being  equal  to  020, 
cb2,  etc.,  and  also  equal  to  the 
distances  ac,  cb,  etc.,  on  the  common  tangent,  it  will  be  clear  that  if  the 
axes  01  and  o2  are  fixed,  the  spirals  may  turn,  fulfilling  the  conditions  of 
perfect  rolling  contact;  for  the  arc  cbi  =  arc  c62,  and  also  the  radiant 
Oibi  +  radiant  o262  =  o/o  +  02'o  =o\c  +  ozc;  and  similarly  for  successive 
arcs  and  radiants. 

;     98.   To  construct  two  spirals,  as  in  Fig.  101,  with  a  given  obliquity.  — 
The  equation    for  such  a  logarithmic 
spiral  is 


and  6 


FIG.  101 


where  a  is  the  value  of  r  when  6  is  zero; 

-    —  /  <£    being    the    constant 
tan  ( 


FIG.  102 


angle  between  the  tangent  to  the  curve 
and  the  radiant  to  the  point  of 
tangency;  and  where  e  is  the  base  of  the  Naperian  logarithms. 
In  Fig.  102  let  oc  =  a,  and  ocd  =  <£.  Taking  successive  values  of  9, 
starting  from  oc,  we  may  calculate  the  values  of  r  and  thus  plot  the 
curve.  If,  however,  it  is  desired  to  pass  a  spiral  through  two  points  on 
radiants  a  given  angle  apart,  it  is  to  be  noticed  from  the  equation  of  the 


TRANSMISSION  OF   MOTION 


79 


curve  that  if  the  successive  values  of  6  are  taken  with  a  uniform  increase, 
the  lengths  of  the  corresponding  radiants  will  be  in  geometrical  progres- 
sion. To  draw  a  spiral  through  the  points  b  and  e,  Fig.  102,  bisect  the 
angle  boe,  and  make  of  a  mean  proportional  between  ob  and  oe;  f  will  be 
a  point  on  the  spiral.  Then  by  the  same  method  bisect  foe,  and  find 
oh;  also  bisect  bof  and  find  o/c,  and  so  on;  a  smooth  curve  through  the 
points  thus  found  will  be  the  desired  spiral. 

99.  Continuous  Motion.  —  Since  these  curves  are  not  closed,  one  pair 
cannot  be  used  for  continuous  motion;  but  a  pair  of  such  curves  may  be 
well  adapted  to  sectional  wheels  requiring  a  varying  angular  speed. 
For  example,  in  Fig.  103,  given  the  axes  o\  and  o2,  the  angle  co\e  through 

which  A  is  to  turn,  and  the  limits  of  the  angular  speed  ratio.     Make  -] 

o2c 

equal  to  the  minimum  angular  speed  ratio  and  — -,  equal  to  the  maximum 

angular  speed  ratio.  Then  Oie  must  equal  Old.  Now  construct  a  spiral 
through  the  points  c  and  e.  The  spiral  for  B  is  that  part  of  the  spiral 


FIG.  103 


FIG.  104 


A  constructed  about  01  which  would  be  included  between  radiants  Oib 
and  Oia,  equal  respectively  to  o2c  and  o2g  (=  o2d),  which  may  be  found 
by  continuing  the  spiral  about  01  beyond  c  or  e  if  necessary.  Since  these 
curves  (Fig.  103)  are  parts  of  the  same  spiral,  and  since  by  construction 
Oic  +  o2c  =  Oie  +  o2g,  A  could  drive  B,  the  points  e  and  g  ultimately  rolling 
together  at  d  on  the  line  of  centers.  The  conditions  of  rolling  contact  are 
evidently  fulfilled,  as  will  be  seen  by  referring  to  §  97. 

100.  Logarithmic  Spiral  Driving  Slide.  —  Fig.  104  shows  a  logarith- 
mic spiral  sector  A  driving  a  slide  B.  Here  the  driven  surface  of  the 
slide  coincides  with  the  tangent  to  the  spiral,  the  line  of  centers  being 
from  o  through  c  to  infinity  and  perpendicular  to  the  direction  of  motion 
of  the  slide.  In  this  combination  the  linear  speed  of  the  slide  will  equal 


80 


ELEMENTS  OF  MECHANISM 


the  angular  speed  of  A  multiplied  by  the  length  of  the  radiant  in  contact 
oc. 

101.  Wheels  using  Logarithmic  Spirals  arranged  to  allow  Complete 
Rotations.  —  By  combining  two  sectors  from  the  same  or  from  dif- 
ferent spirals,  unilobed  wheels  may  be  found  which  may  be  paired  in 
such  a  way  as  to  fulfill  the  laws  of  perfect  rolling  contact.  Taking  two 
equal  sectors  from  the  same  spiral,  we  should 
have  a  symmetrical  unilobed  wheel,  as  A  (Fig. 
105),  and  this  will  run  perfectly  with  a  wheel  B 
exactly  like  A,  as  shown.  If  A  is  the  driver,  the 
minimum  angular  speed  of  B  will  occur  when  the 
points  d  and  e  are  in  contact,  and  we  shall  have 

a.v.  B  _  Old 
a.v.  A      026 

The  maximum  angular  speed  of  B  will  occur  when 
the  points  /  and  g  are  in  contact.  Such  wheels  are 
readily  formed,  if  the  maximum  and  minimum 
angular  speed  ratios  are  known,  by  the  method  in 
§  97,  only  it  is  to  be  noticed  that  the  minimum 
ratio  must  be  the  reciprocal  of  the  maximum 
ratio,  and  that  the  angle  which  each  sector 
subtends  must  be  180°  Unilobed  wheels  need  not  be  formed  from 
equal  sectors,  in  which  case  the  sectors  used  will  not  have  the  same 
obliquity  nor  will  the  subtended  angles  be  equal,  but  the  wheels  must  be 
so  paired  that  sectors  of  the  same  obliquity  shall  be 
in  contact.  Fig.  106  shows  a  pair  of  such  wheels 
in  which  maximum  and  minimum  angular  speed 
ratios  occur  at  unequal  intervals;  it  will,  however, 
be  noticed  that  the  minimum  angular  speed  ratio 
must  here  also  be  the  reciprocal  of,  the  maximum 
ratio. 

By  a  similar  method  wheels  may  be  formed  which 
shall  give  more  than  one  position  of  maximum  and  of 
minimum  angular  speed  ratio;  that  is,  there  may  be 
either  symmetrical  or  unsymmetrical  bilobed  wheels, 
trilobed  wheels,  etc.  Fig.  107  shows  a  pair  of  sym- 
metrical bilobed  wheels.  Here  all  the  sectors  are  from 
the  same  spiral,  all  the  same  length,  each  subtending 
an  angle  of  90°.  It  will  be  seen  that  the  conditions  of  rolling  contact 
are  perfectly  fulfilled  and  that  if  A  turns  uniformly  B  will  have  two 
positions  of  maximum  and  two  of  minimum  speed.  Similarly  a 


FIG.  105 


FIG.  106 


TRANSMISSION   OF  MOTION 


81 


pair  of  symmetrical  trilobed  wheels  could  be  formed  where  each  of  the 
sectors  subtends  an  angle  of  60°. 

Following  the  method  used  in  obtaining  the  unsymmetrical  unilobed 
wheels  of  Fig.  106,  a  pair  of  unsymmetrical  bilobed  wheels  could  be 

arranged,    provided  only  that 

sectors  of  the   same  obliquity 

come   into   contact    and    that 

such    sectors     subtend     equal 

angles.     Fig.  108  shows  a  pair 

of  trilobed  wheels  of  this  form. 
Such   wheels   as   those  just 

described  cannot  be  inter- 
changeable, but  since  any  two 

spiral    arcs   having    the    same 

obliquity  will  roll  correctly,  a 

unilobe  may  be  made  to  roll 

correctly  with  a  bilobe  where 

the  sectors  of  the  unilobe  are 

a  given  spiral  and  each  subtending  180°,  and  where  each  of 
the  sectors  of  the  bilobe  is  of  the  same  length  as  one  of  those  of  the 
unilobe,  and  from  a  spiral  of  the  same  obliquity,  but  where  each 
subtends  an  angle  of  90°.  In  a  similar  manner  a  trilobed  wheel  may  be 
found  which  could  be  driven  by  the  same  unilobed  wheel  as  above,  hence 


FIG.  107 


FIG.  108 


from 


FIG.  109 

also  by  the  bilobed  wheel  found  from  that  unilobe.  These  wheels  would 
therefore  be  interchangeable.  Fig.  109  shows  a  set  of  such  wheels  which 
would  be  symmetrical  wheels.  A  set  of  unsymmetrical  wheels  could 
be  found  in  a  similar  manner. 

102.   The  Rolling  of  Equal  Ellipses.  —  If  two  equal  ellipses,  each 
turning  about  one  of  its  foci,  are  placed  in  contact  in  such  a  way  that  the 


82 


ELEMENTS  OF  MECHANISM 


distance  between  the  axes  Oi02,  Fig.  110,  is  equal  to  the  major  axis  of 
the  ellipses,  we  shall  find  that  they  will  be  in  contact  on  the  line  of  centers 
and  that  the  rolling  arcs  are  of  equal  length.  If  the  point  c  is  on  the  line 
of  centers  Oio2,  we  should  have  QIC  -f-  co2  =  Oio2  =  OiC  +  cd,  and  therefore 
cd  =  co2.  Since  the  tangent  to  an  ellipse  at  any  point,  as  c,  makes  equal 
angles  with  the  radii  from  the  two  foci,  Oicm  =  den  and  ecm  =  o2cn;  but 

since  cd  =  co2,  the  point  c  is  simi- 
larly situated  in  the  two  ellipses,  and 
therefore  the  angle  Oicm  would  equal 
the  angle  o2cn,  which  would  give  a 
common  tangent  to  the  two  curves 
at  c.  Hence  if  Oi02  is  equal  to  the 
major  axis,  the  ellipses  could  be  in 
rolling  contact  on  the  line  Oi02. 
Since  the  distances  cd  and  co2,  from 
the  foci  d  and  o2  respectively,  are 
equal,  it  also  follows  that  the  arc 
pIG  HQ  cf  is  equal  to  the  arc  eg  which  com- 

pletes the  requirements  for  perfect 

rolling  contact.  It  will  also  be  noticed  that  the  line  dee  will  be  straight 
and  that  a  link  could  connect  d  and  e}  as  will  be  seen  when  discussing 
linkwork. 

If  A  (Fig.  110)  is  the  driver,  the  angular  speed  ratio  will  vary  from  a 

minimum  when  h  and  k  are  in  contact,  and  then  equal  to  ^y  j  to  a  maxi- 

02/C 

mum  when  /  and  g  are  in  contact,  when  it  will  equal  —  •    The  angular 

02  g 

speed  ratio  will  be  unity  when  the  major  axes  are  parallel,  the  point  of 
contact  being  then  midway  between  01  and  o2 . 

Such  rolling  ellipses  supplied  with  teeth,  thus  forming  elliptic  gears, 
are  sometimes  used  to  secure  a  quick-return  motion  in  a  slotting 
machine. 

103.  Multilobed  Wheels  Formed  from  Rolling  Ellipses.  —  Non- 
interchangeable  wheels  may  be  formed  from  a  pair  of  ellipses  by  con- 
tracting the  angles  the  same  amount  in  each  ellipse.  Thus,  if  the  angles 
were  contracted  to  one-half  their  size,  a  pair  of  bilobed  wheels  could  be 
formed;  and  if  to  one-third  their  size,  a  pair  of  trilobed  wheels.  Such 
wheels  would  give  perfect  rolling  contact,  but  could  only  be  used  in  pairs 
as  stated. 

By  a  different  method  of  contraction  a  pair  of  wheels  may  be  formed, 
one  of  which  may  be,  for  example,  a  bilobe  and  the  other  a  trilobe.  By 
this  method  only  parts  of  the  original  ellipses  are  used;  parts  which 


TRANSMISSION  OF   MOTION 


83 


would  roll  correctly,  but  which  subtend  unequal  angles  in  some  desired 
ratio.  If  the  arcs  subtend  angles  in  proportion  as  2  is  to  3,  the  angles 
may  be  contracted  or  expanded  to  be  60°  and  90°,  which  are  in  the  same 
ratio,  when  we  shall  have  arcs  suitable  for  a  trilobe  and  a  bilobe  respec- 
tively, which  will  roll  correctly.  For  example,  assume  the  foci  01  and  d 
(Fig.  Ill) ;  lay  off  angles  foid  and  fde  as  2  to  3.  Then  the  point  /  will 
lie  on  an  ellipse  from  which  a  bilobe  and 
a  trilobe  may  be  formed  by  contracting 
the  angle  foid  to  60°  and  the  angle  go2d 
=  fde  to  90°,  as  shown  in  the  figure. 

104.  The  Rolling  of  Equal  Parabolas. 
-  Two  parabolas  may  be  considered  as 

two  ellipses  with  one  focus  of  each  re- 
moved to  infinity.  In  the  ellipses  of  Fig. 
110  suppose  the  foci  01  and  e  to  be  so 
removed;  we  shall  have  the rparabolas  of 
Fig.  112  in  contact  at  the  point  c  and  in 
perfect  rolling  contact,  one  turning  about 
its  focus  02  as  an  axis,  and  the  other 
having  a  motion  of  translation  perpen- 
dicular to  Old. 

To  prove  the  rolling  action  perfect, 
assume  the  parabolas  with  their  vertexes 
in  contact  at  m.  Let  /  be  the  point  on 
the  turning  parabola  which  will  move  to  c, 
so  that  o2/  =  o2c.  Draw  fg  parallel  to  o2c,  and  since  the  parabolas  are 
equal  we  shall  have  Ig  =  o%  /,  therefore  Ig  =  o2c;  but  since  ozk  is  the 
directrix  of  the  parabola  whose  focus  is  now  at  I,  Ig  =  gk;  therefore 
gk  =  Ozc,  and  as  this  parabola  slides  perpendicular  to  o\d,  the  point  g 
would  also  move  to  c.  The  rolling  arcs  mf  and  mg  are  equal.  Thus 
the  parabola  turning  about  o2  would  cause  the  other  parabola  to  have 
translation  perpendicular  to  Old,  the  two  moving  in  perfect  rolling 
contact. 

105.  The  Rolling  of  Equal  Hyperbolas.  —  If  two  equal  hyperbolas 
are  placed,  as  in  Fig.  113,  so  that  the  distances  between  their  foci  01  and 
02,  and  d  and  e,  are  each  equal  to  fg  =  hk,  the  distance  between  the  ver- 
texes of  the  hyperbolas,  we  shall  find  them  in  contact  at  some  point  c. 
If  the  foci  0i  and  o2  are  then  taken  as  axes  of  rotation,  the  hyperbolas  will 
turn  in  perfect  rolling  contact.     To  prove  this  take  the  point  I  on  the 
hyperbola  whose  foci  are  at  0i  and  d  so  that  oj,  =  dc  and  QIC  =  dl.     Then 
since  a  tangent  at  any  point  on  a  hyperbola  makes  equal  angles  with 
the  radii  from  the  two  foci,  the  tangent  at  I  will  bisect  the  angle  ojd 


FIG.  Ill 


84 


ELEMENTS  OF  MECHANISM 


and  the  tangent  at  c  will  bisect  the  equal  angle  Oicd.  If  now  the  branch 
Oihl  is  placed  tangent  to  the  branch  dkc  with  the  points  I  and  c  in  con- 
tact, the  radius  loi  must  fall  on  Ojc  and  dl  on  dc.  Since  the  difference 


e  atlrrft/tity 


FIG.  112 


FIG.  113 

between  the  radii  from  the  two  foci  to  any  point  on  a  hyperbola  is  a 
constant  and  equal  to  the  distance  between  the  vertexes,  QIC  —  dc  =  hk; 
but  oil  was  taken  equal  to  dc,  hence  QIC  —  oj,  =  hk.  Then,  since  Oi02  was 


TRANSMISSION  OF  MOTION  85 

originally  assumed  equal  to  hk,  we  shall  have  QIC  —  oj  =  Oi02,  and  there- 
fore the  line  0i02c  will  be  a  straight  line,  and  the  point  of  contact  c  will 
lie  on  the  line  of  centers.  The  arc  Ih  which  is  equal  to  ck  will  also  be 
equal  to  cf.  Therefore  the  hyperbolas  will  be  in  perfect  rolling  contact. 
The  same  reasoning  will  apply  for  any  position  of  the  point  of  contact. 
It  will  be  seen  in  a  later  chapter  that  since  OiOz  =  de  =  a  constant,  and 
Old  =  o&  —  a  constant,  the  linkage  o^ed  with  the  axes  01  and  o2  fixed 
would  cause  the  same  angular  speed  ratio  about  01  and  o2  as  the  rolling 
hyperbolas  would  give. 

If  the  hyperbola  turning  about  the  axis  o2  is  the  driver,  the  angular 
speed  ratio  will  be  a  minimum  when  the  vertexes  /  and  k  are  in  contact 

and  will  be  — j- ;  this  ratio  will  increase  as  the  point  of  contact  approaches 

infinity,  when  the  ratio  would  be  unity,  and  would  correspond  to  the  posi- 
tion of  the  linkage  when  Oi02  and  de  are  parallel.  Further  rotation  would 
bring  the  opposite  branches  of  the  hyperbolas  into  contact,  the  maxi- 
mum angular  speed  ratio  occurring  as  the  points  g  and  h  come  together, 

QnQ 

when  its  value  becomes  -~  •  The  construction  shown  in  the  figure  will 
allow  only  a  limited  motion. 


CHAPTER  V 
GEARS  AND   GEAR  TEETH 

106.  Gear  Drives.     It  was  shown  in  Chapter  IV  that  one  shaft 
could  cause  another  to  turn  by  means  of  two  bodies  in  pure  rolling 
contact.     If  the  speed  ratio  must  be  exact  or  if  much  power  is  to  be 
transmitted,  a  drive  depending  solely  upon  friction  between  the  sur- 
faces of  the  rolling  bodies  is  not  sufficiently  positive.     For  this  reason 
toothed  wheels,  called  gears,  are  used  in  place  of  the  rolling  bodies. 
As  the  gears  turn  the  teeth  of  one  gear  slide  on  the  teeth  of  the  other 
but  are  so  designed  that  the  angular  speeds  of  the  gears  are  the  same 
as  those  of  the  rolling  bodies  which  they  replace. 

107.  Gearing  Classified.     In  §  83  attention  was  called  to  the  fact 
that  rolling  bodies  may  be  used  to  connect  axes  which  are  parallel, 
intersecting,   or  neither  parallel  nor  intersecting.     The  same  cases 
arise  in  the  use  of  gears,  and  special  names  are  given  to  the  gears 
according  to  the  case  for  which  they  are  designed. 

Gears  may  be  classified  on  the  above  basis  as  follows: 

External  Gears  —  Fig.  114. 
Internal  Gears  —  Fig.  115 
(Here  the  large  gear  is  called  an  annular 

and  the  small  one  a  pinion)  Connecting 

Spur   Gears    Twisted  Spur  Gear  —  Fig.  116.  Parallel 

Herring  Bone  Spur  Gear  —  Fig.  117.  Axes 

Rack  and  Pinion  —  Fig.  118. 

(The  rack  is  a  gear  of  infinite  radius) 
Pin  Gearing  —  Fig.  119. 
Plain  Bevel  (including  Mitre  Gears,  which 

are  equal  bevel  gears  on  shafts  at  90°)  -      Connecting 
Bevel  Gears  I      Fig.  120.  Intersect- 

Crown  Gears  —  Fig.  121.  ing  Axes 

Twisted  Bevel  Gears  —  Fig.  122. 

Hyperboloidal  or  Skew  Gears-Fig.  123.  {  Connecting  Axes  in  differ- 

i      ent  pianes. 

Srrpw  Goring-  I     Worm  and  Wheel  —  Fig.  124.   }  Connecting  Axes  in 
ng  1    Helical  Gears  —  Fig.  125.          J      different  planes. 

The  name  pinion  is  often  applied  to  the  smaller  of  a  pair  of  gears. 
The  various  kinds  of  gears  enumerated  above  will  be  discussed  in 
more  detail  after  the  principles  which  apply  to  gearing  in  general 

have  been  considered. 

86 


GEARS  AND  GEAR  TEETH 


87 


FIG.  114 


FIG.  115 


88 


ELEMENTS  OF   MECHANISM 


FIG.  116 


FIG.  117 


FIG.  118 


GEARS  AND  GEAR  TEETH 


89 


FIG.  119 


FIG.  120 


90 


ELEMENTS  OF  MECHANISM 


a 


FIG.  121 


Fia.  122 


GEARS  AND  GEAR  TEETH 


91 


FIG.  123 


FIG.  124 


92 


ELEMENTS  OF  MECHANISM 


108.  Speed  Ratio  of  a  Pair  of  Gears.  It  has  been  shown. in  the 
preceding  chapter  that  if  two  cylinders  as  A  and  B,  Fig.  126,  are  keyed 
to  the  shafts  S  and  Si  respectively,  the  angular  speed  of  S  is  to  the 
angular  speed  of  Si  as  Di  is  to  D,  provided  there  is  sufficient  friction 
between  the  circumferences  of  the  discs  to  prevent  one  slipping  on  the 
other.  If  the  speed  ratio  must  be  exact,  or  if  much  power  is  to  be 
transmitted,  a  drive  like  this,  depending  solely  upon  friction,  is  not 

positive  enough.  To  make 
sure  that  there  shall  be 
no  slipping,  wheels  having 
teeth  around  their  circum- 
ferences are  substituted 
for  the  plain  discs.  The 
outlines  of  these  teeth 
must  be  such  that  the 
speed  ratio  is  constant. 
Such  a  pair  of  wheels  is 
shown  in  Fig.  127.  Here 
the  larger  gear  has  16  teeth 
and  the  smaller  gear  12 
teeth.  Assume  that  the 
shaft  S  is  being  turned 
from  some  external  source 
of  power;  the  gear  A, 
since  it  is  keyed  to  S,  will 
turn  with  it.  Then  the 
teeth  on  A  will  push  the 
teeth  on  B,  a  tooth  on  A 
coming  in  contact  with  a 
tooth  on  B  and  pushing 
/  that  tooth  along  until  tfie 
gears  have  turned  so  far  around  that  tbo^k  two  teeth  swing  out  of  reach 
of  each  other.  In  order  for  B  to  make  a  complete  revolution  each  one 
of  its  12  teeth  must  be  pushed  along  thus  past  the  center  line.  There- 
fore, while  B  turns  once  12  of  the  teeth  on  A  must  pass  the  center  line 
Since  A  has  16  teeth  in  all,  A  will  therefore  make  {$  of  a  turn  while  B 
makes  one  turn.  In  other  words,  the  turns  of  A  in  a  given  time  are  to 
the  turns  of  B  in  the  same  time  as  the  number  of  teeth  on  B  are  to  the 
number  of  teeth  on  A. 

It  is  evident  that  the  distance  from  the  center  of  one  tooth  to  the 
center  of  the  next  tooth  on  both  gears  must  be  alike  in  order  that 
the  teeth  on  one  may  mesh  into  the  spaces  on  the  other^ 


FIG.  126 


GEARS  AND  GEAR  TEETH 


93 


FIG.  126 


Addendum 

loot  or  Dedendum 

FIG,  127 


/  Tooth  Face 
Tooth  Flank 


94  ELEMENTS  OF  MECHANISM 

109.  Pitch  Circles  and  Pitch  Point.     Let  a  point  P  (Fig.  127)  be 
found  on  the  center  line  SSi  such  that  ^-g-  =         , and  through 

this  point  draw  circles  about  S  and  Si  as  centers.  Call  their  diameters 
D  and  Di.  Then  D  =  2  P/S  and  DI  =  2  P&.  Since,  as  shown  above, 

Revolutions  of  B  _  Teeth  on  A 
Revolutions  of  A      Teeth  on  B ' 
therefore, 

Revolutions  of  B  _  D^ 
Revolutions  of  A      DI 

That  is,  the  two  gears  when  turning  will  have  the  same  speed  ratio 
as  would  two  rolling  cylinders  of  diameters  D  and  DI.  The  point  P 
which  divides  the  line  of  centers  of  a  pair  of  gears  into  two  parts  pro- 
portional to  the  number  of  teeth  in  the  gears  is  called  the  pitch  point. 
The  circle  D,  drawn  through  P  with  center  at  S,  is  the  pitch  circle  of 
the  gear  A  and  the  circle  DI  is  the  pitch  circle  of  the  gear  B. 

110.  Addendum  and  Root  Circles.     The  circle  passing  through 
the  outer  ends  of  the  teeth  of  a  gear  is  called  the  addendum  circle 
and  the  circle  passing  through  the  bottom  of  the  spaces  is  called  the 
root  circle. 

111.  Addendum  Distance  and  Root  Distance.    Length  of  Tooth. 
The  radius  of  the  addendum  circle  minus  the  radius  of  the  pitch 
circle  is  the  addendum  distance,  or,  more  commonly,  the  addendum. 
The  radius  of  the  pitch  circle  minus  the  radius  of  the  root  circle  is  the 
root  distance  or  root  or  dedendum.     The  root  plus  the  addendum  is 
the  length  of  tooth. 

112.  Face  and  Flank  of  Tooth.    Acting  Flank.     That  portion  of 
the  tooth  curve  which  is  outside  the  pitch  circle  is  called  the  face  of  the 
tooth  or  tooth  face.     This  must  not  be  confused  with  the  term  "  face 
of  gear"  (§  113).     The  part  of  the  tooth  curve  inside  the  pitch  circle 
is  called  the  flank  of  the  tooth. 

That  part  of  the  flank  which  comes  in  contact  with  the  face  of  the 
tooth  of  the  other  gear  is  called  the  acting  flank. 

113.  Face  of  Gear.     The  length  of  the  gear  tooth  measured  along 
an  element  of  the  pitch  surface  is  called  the  length  of  the  face  of  the 
gear  or  width  of  face  of  the  gear.     (See  top  view,  Fig.  127.) 

114.  Clearance.     The  distance  measured  on  the  line  of  centers, 
between  the  addendum  circle  of  one  gear  and  the  root  circle  of  the 
other,  when  they  are  in  mesh,  is  the  clearance. 

This  is  evidently  equal  to  the  root  of  one  gear  minus  the  addendum 
of  the  mating  gear. 


GEARS  AND  GEAR  TEETH  95 

116.  Backlash.  When  the  width  of  a  tooth,  measured  on  the  pitch 
circle,  is  less  than  the  width  of  the  space  of  the  gear  with  which  it  is 
in  mesh,  the  difference  between  the  width  of  space  and  width  of  tooth 
is  called  the  backlash.  This  is  shown  in  Fig.  128  where  S  minus  T  is 
the  backlash,  S  and  T  being  understood  to  be 
measured  on  the  arcs  of  the  pitch  circles.  Accu- 
rately made  gears  rarely  have  any  appreciable 
amount  of  backlash,  but  cast  gears  or  roughly 
made  gears  require  backlash. 

116.  Circular  Pitch.     The  distance  from  the 
center  of  one  tooth  to  the  center  of  the  next 
tooth,   measured  on  the  pitch  circle,  is  called 
the  circular  pitch.     This  is,  of  course,  equal  to 
the  distance  from  any  point  on  a  tooth  to  the 

T^T/"1        1  9& 

corresponding  point  on  the  next  tooth  meas- 
sured  along  the  pitch  circle.  (See  Fig.  127.)  The  circular  pitch  is 
equal  to  the  width  of  tooth  plus  the  width  of  the  space  between  teeth,  meas- 
ured on  the  pitch  circle.  The  whole  circumference  of  the  pitch  circle 
is  equal  to  the  circular  pitch  multiplied  by  the  number  of  teeth,  or 
the  circular  pitch  is  equal  to  the  circumference  of  the  pitch  circle 
divided  by  the  number  of  teeth.  In  Fig.  127  let  T  represent  the 
number  of  teeth  in  the  gear  A  and  let  C  represent  the  circular  pitch. 
Then, 

C  =  "~£-  (44) 

Two   gears   which    mesh    together    must    have   the    same   circular 
pitch. 

117.  Diametral  Pitch  and  Pitch  Number.     Module.     The  term 
diametral  pitch  is  used  by  different  authorities  to  mean  two  different 
quantities.     Most  gear  makers'  catalogues  and  many  books  on  gearing 
define  diametral  pitch  as  the  number  of  teeth  per  inch  of  diameter  of 
pitch  circle.     For  example,  if  a  gear  has  24  teeth  and  the  diameter  of 
its  pitch  circle  is  8  in.,  these  authorities  would  say  that  the  diametral 
pitch  of  the  gear  is  24  divided  by  8  or  is  3.     Such  a  gear  is  described 
as  a  3-pitch  gear.     This  is  also  called  the  pitch  number. 

Other  authorities  define  diametral  pitch  as  the  length  of  pitch 
diameter  which  the  gear  has  per  tooth,  or  the  ratio  pf  the  diameter 
to  the  number  of  teeth.  That  is,  referring  again  to  the  24  tooth  gear 
having  a  pitch  diameter  of  8  in.,  according  to  this  latter  definition  the 
diametral  pitch  would  be  8  in.  divided  by  24  or  J  of  an  inch.  Another 
name  for  this  quantity  is  module. 


96  ELEMENTS  OF  MECHANISM 

Throughout  this  book  the  terms  diametral  pitch  and  pitch  number 
will  be  used  interchangeably,  meaning  the  number  of  teeth  per  inch 
of  pitch  diameter,  and  the  name  module  will  be  used  for  the  amount 
of  diameter  per  tooth. 

If  M  represents  the  module  and  P.N.  the  pitch  number  or  diametral 
pitch,  T  the  number  of  teeth  and  D  the  pitch  diameter,  the  above 
may  be  expressed  in  the  form  of  equations  as  follows: 

,..      Pitch  diameter      D 

M-     --     =T' 


p  \T    _  Teeth 


Pitch  diameter      D 
Therefore,  M  =  p  „  • 

118.  Relation  between  Circular  Pitch  and  Module. 
From  Eq.  (45)  M  =  ^> 

and  from  Eq.  (44)  C  =  ^  • 

Dividing  Eq.  (44)  by  Eq.  (45), 

C_  _  TT!)  ^  D  _ 
M  ==   T  "'   T  " 
or  C  =  If  X  T,  (47) 

or  C  =  p;k-  <48) 

Or,  in  words,  the  circular  pitch  is  equal  to  the  module  multiplied  by  TT. 

119.  Angle  and  Arc  of  Action.     The  angle  through  which  the  driv- 
ing gear  turns  while  a  given  tooth  on  the  driving  gear  is  pushing  the 
corresponding  tooth  on  the  driven  gear  is  called  the  angle  of  action 
of  the  driver.     Similarly,  the  angle  through  which  the  driven  gear 
turns  while  a  given  one  of  its  teeth  is  being  pushed  along  is  called  the 
angle  of  action  of  the  driven  gear.     The  angle  of  approach,  in  each 
case,  is  the  angle  through  which  the  gear  turns  from  the  time  a  pair 
of  teeth  come  into  contact  until  they  are  in  contact  at  the  pitch  point. 
It  will  be  shown  later  that  the  pitch  point  is  one  of  the  points  of  con- 
tact of  a  pair  of  teeth  during  the  action.     The  angle  of  recess  is  the 
angle  turned  through  from  the  time  of  pitch  point  contact  until  con- 
tact ceases. 

The  angle  of  action  is  therefore  equal  to  the  angle  of  approach, 
plus  the  angle  of  recess. 


GEARS  AND  GEAR  TEETH  97 

In  Fig.  129  a  tooth  M  on  the  driving  gear  is  shown  (in  full  lines)  just 
beginning  to  push  a  tooth  N  on  the  driven  gear.  The  dotted  lines 
show  the  position  of  the  same  pair  of  teeth  when  N  is  just  swinging  out 
of  reach  of  M.  While  M  has  been  pushing  N,  any  radial  line  on  the 
gear  B,  as,  for  example,  the  line  drawn  through  the  center  of  the  tooth 
M,  has  swung  through  the  angle  K,  and  any  line  on  gear  A  has  swung 
through  the  angle  V.  K  is,  there- 
fore, the  angle  of  action  of  the  gear 

B,  and  V  is  the  angle  of  action  of  _^--^W*-7"X.  Drlyer 

the  gear  A. 

It  should  be  noted  that  the 
angles  of  approach  and  recess  are 
not  shown  in  Fig.  129. 

The  arc  of  action  is  the  arc  of  the 
pitch  circle  which  subtends  its  angle  of  action.  The  arcs  of  approach 
and  recess  bear  the  same  relation  to  the  angles  of  approach  and  recess 
as  the  arc  of  action  bears  to  the  angle  of  action.  Since  the  arcs  of 
action  on  both  gears  must  be  equal,  the  angles  of  action  must  be 
inversely  as  the  radii. 

Therefore  the  following  equation  holds  true: 

Angle  of  action  of  driver      _  Number  of  teeth  on  driven  gear      ( .  , 
Angle  of  action  of  driven  gear          Number  of  teeth  on  driver 

The  arc  of  action  must  never  be  less  than  the  circular  pitch,  for,  if  it 
were,  one  pair  of  teeth  would  cease  contact  before  the  next  pair  came 
into  contact. 

120.  The  Path  of  Contact.     Referring  still  to  Fig.  129,  the  teeth 
as  shown  in  full  lines  are  touching  each  other  at  one  point  a.     This 
point  is  really  the  projection  on  the  plane  of  the  paper  of  a  line  of  con- 
tact equal  in  length  to  the  width  of  the  gear  face  (see  §  113).     In  the 
position  shown  dotted  the  teeth  touch  each  other  at  the  point  b.     If 
the  teeth  were  drawn  in  some  intermediate  position,  they  would  touch 
at  some  other  point.     For  every  different  position  which  the  teeth 
occupy  during  the  action  of  one  pair  of  teeth  they  have  a  different 
point  of  contact.     A  line  drawn  through  all  the  points  at  which  the 
teeth  touch  each  other  (in  this  case  the  line  aPb)  is  called  the  path  of 
contact.     This  may  be  a  straight  line  or  a  curved  line,  depending  upon 
the  nature  of  the  curves  which  form  the  tooth  outlines.    In  all  properly 
constructed  gears  the  pitch  point  P  is  one  point  on  the  path  of  contact. 

121.  Obliquity  of  Action  or  Pressure  Angle.     The  angle  between  the 
line  drawn  through  the  pitch  point  perpendicular  to  the  line  of  centers, 
and  the  line  drawn  from  the  pitch  point  to  the  point  where  a  pair  of 


98 


ELEMENTS  OF  MECHANISM 


teeth  are  in  contact  is  called  the  angle  of  obliquity  of  action  or  pressure 
angle.  In  some  forms  of  gear  teeth  this  angle  remains  constant  while 
in  other  forms  of  teeth  it  varies. 

The  direction  of  the  force  which  the  driving  tooth  exerts  on  the 
driven  tooth  is  along  the  line  drawn  from  the  pitch  point  to  the  point 
where  a  pair  of  teeth  are  in  contact  (see  §  122).  The  smaller  the 
angle  of  obliquity  the  greater  will  be  the  component  of  the  force  in 
the  direction  to  cause  the  driven  gear  to  turn  and  the  less  will  be  the 
tendency  to  force  the  shafts  apart.  In  other  words,  a  large  angle  of 
obliquity  tends  to  produce  a  large  pressure  on  the  bearings. 

122.  Law  Governing  the  Shape  of  the  Teeth.  The  curves  which 
form  the  outline  of  the  teeth  on  a  pair  of  gears  may,  in  theory  at  least, 
have  any  form  whatever,  provided  they  conform  to  one  law,  namely: 
The  line  drawn  from  the  pitch  point  to  the  point  where  the  teeth  are  in 
contact  must  be  perpendicular  to  a  line  drawn  through  the  point  of  con- 
tact tangent  to  the  curves  of  the  teeth. 

That  is,  the  common  normal  to  the  tooth  curves  at  all  points  of  contact 
must  pass  through  the  pitch  point. 

This  is  illustrated  in  Fig.  130.  The  teeth  in  the  full  line  position 
touch  each  other  at  a.  That  is,  the  curves  are  tangent  to  each  other 


\ 


at  this  point.  The  line  ST  is  drawn  tangent  to  the  two  curves  at  a. 
The  curves  must  be  made  so  that  this  tangent  line  is  perpendicular 
to  the  line  drawn  from  a  to  P.  Similarly,  in  the  dotted  position  the 
line  VW  which  is  tangent  to  the  curves  at  their  point  of  contact  b 
must  be  perpendicular  to  the  line  b  P.  This  must  hold  true  for  all 
positions  in  which  a  pair  of  teeth  are  in  contact,  in  order  that  the 
speed  ratio  of  the  gears  shall  be  constant. 

This  law  may  be  proved  as  follows:   Let  An  and  Bm  be  lines  drawn 
from  the  centers  A  and  B  perpendicular  to  the  common  normal  through 


GEARS  AND  GEAR  TEETH  99 

the  point  of  contact  a.  Let  aA  =  angular  speed  of  gear  A  (expressed 
in  radians)  and  WB  =  the  angular  speed  of  B.  Then  linear  speed  of 
n  =  COA  X  An  and  linear  speed  of  m  =  co*  X  Bm.  The  direction  of 
motion  of  m  and  n  are  both  along  nm  at  the  instant  under  consideration 
and  the  motion  is  the  same  as  if  wheel  A  were  pulling  B  by  an  inexten- 
sible  cord  attached  at  m  and  n.  That  is,  the  linear  speed  of  m  =  linear 

speed  of  n.    Therefore  w4  X  An  =  co*  X  Bm  or^  =  ^--     But  since 

the  angular  speeds  of  the  radii  to  two  points  which  have  the  same  linear 

,..   COA       BP 
speed  are  inversely  as  the  radii,  —  =  jp- 


Bm      BP 
Therefore,  JZ  =  AP 

Hence  nm  must  intersect  the  line  of  centers  AB  at  P. 

123.  Conjugate  Curves.     Two  curves  are  said  to  be  conjugate 
when  they  are  so  formed  that  they  may  be  used  for  the  outline  of 
two  gear  teeth  which  will  work  on  each  other  and  fulfill  the  law 
described  in  §  122. 

124.  To  Draw  a  Tooth  Outline  which  shall  be  Conjugate  to  a  Given 
Tooth  Outline.     Given  the  face  or  flank  of  a  tooth  of  one  of  a  pair  of 
wheels,  to  find  the  flank  or  face  of  a  tooth  of  the  other.     The  solution 
of  this  problem  depends  on  the  fundamental  law,  §  122.     In  Fig.  131 
let  the  flank  and  face  of  a  tooth  on  A  be  given.     If  A  is  considered  as 
the  driver,  points  on  the  flank,  as  a  and  6,  will  be  points  of  contact  in 
the  approaching  action,  and  by  the  law  they  can  properly  be  points 
of  contact  only  when  the  normals  to  the  flank  at  these  points  pass 
through  the  pitch  point;  therefore  drawing  ac  and  bd  normals  to  the 
flank  from  the  points  a  and  b  respectively,  and  then  turning  A  back- 
ward until  the  points  c  and  d  are  at  the  pitch  point,  we  find  positions 
ai  and  61  which  a  and  b  respectively  must  occupy  when  they  can  be 
points  of  contact  with  the  face  of  a  tooth  of  the  other  wheel.     The 
point  ai  must  be  a  point  on  the  desired  face  of  a  tooth  on  the  wheel  B 
when  the  pitch  circles  have  been  moved  backward  an  arc  equal  to 
dc,  that  is,  so  that  c  is  at  the  pitch  point.     To  find  this  point  when 
the  teeth  are  in  the  original  position,  it  is  necessary  to  move  the  wheels 
forward,  the  wheel  B  carrying  with  it  the  point  a\  and  the  normal 
aid  until  the  point  c\  has  moved  through  an  arc  C\GZ  equal  to  Cic;  this 
will  carry  the  point  a\  to  0%,  and  the  normal  aid  to  0202.     During  this 
same  forward  motion  the  normal  a\d  moving  with  the  wheel  A  will 
return  to  its  original  position  ac. 

In  a  similar  manner  the  point  61,  which  can  be  a  point  of  contact  of 


100 


ELEMENTS  OF  MECHANISM 

0,0 


8 


02 

FIG.  131 


GEARS  AND 


101 


the  given  flank  with  the  desired  face,  is  a  point  on  this  face  when  the 
pitch  circles  of  the  wheels  are  moved  backward  an  arc  equal  to  Cid. 
Moving  them  forward  the  same  distance,  the  point  61  and  normal  6iCi, 
moving  with  the  wheel  B,  will  be  found  at  62  and  b2dz.  This  process 
may  be  continued  for  as  many  points  as  may  be  needed  to  give  a 
smooth  curve.  The  curve  drawn  through  the  points  a^b^Ci  will  be 
the  required  face. 

A  similar  process  gives  the  flank  of  the  tooth  on  the  wheel  B  which 
will  work  properly  with  the  given  face.     The  normals  taken  in  the 


B 


FIG.  132 

figure  are  eg  and  fh,  the  positions  of  e  and  /  when  they  can  be  points 
of  contact  being  e±  and  /ij  and  the  points  on  the  required  flank  when 
in  the  original  position  are  e2  and  /2. 

A  smooth  curve  passed  through  the  points  of  contact  ai&iCiei/i  will 
be  the  path  of  contact,  the  beginning  and  end  of  which  will  be  deter- 
mined by  the  addendum  circles  of  B  and  of  A  respectively. 

Fig.  132  shows  the  same  construction  for  finding  the  curve  for  a 
pinion  tooth  conjugate  to  a  given  rack  tooth. 


102 


ELEMENTS  OF  MECHANISM 


Another  method  of  solving  the  above  problem  is  shown  in  Fig.  133, 
where  01  and  o2  are  a  pair  of  plates  whose  edges  are  shaped  to  arcs  of 
the  given  pitch  circles  AAi  and  BB\,  due  allowance  being  made  for  a 
thin  strip  of  metal,  gh,  connecting  the  plates,  to  insure  no  slipping  of 
their  edges  on  each  other. 

Attach  to  02  a  thin  piece  of  sheet  metal,  M,  the  edge  of  which  is 
shaped  to  the  given  curve  aa;  and  to  01  a  piece  of  paper,  D,  the  piece 
M  being  elevated  above  Oi  to  allow  space  for  the  free  movement  of  D. 

Now  roll  the  plates  together,  keep- 
ing the  metallic  strip  gh  in  tension, 
and,  with  a  fine  marking-point, 
trace  upon  the  paper  D,  for  a  suffi- 
cient number  of  positions,  the  out- 
line of  the  curve  aai.  A  curve  just 
touching  all  the  successive  outlines 
on  D,  as  eei,  is  the  corresponding 
tooth  curve  for  o\. 

125.  To  Draw  the  Teeth  of  a 
Pair  of  Gears.  When  the  tooth 
outlines  have  been  found,  and 
the  circular  pitch,  backlash,  ad- 
dendum, and  clearance  are  known, 
the  teeth  may  be  drawn  as 
shown  in  Fig.  134.  Let  MN  and 
RT  be  the  known  tooth  outlines 
for  the  gears  A  and  B  respectively. 
To  draw  three  teeth  on  each  gear,  one  pair  of  which  shall  be 
in  contact  at  the  pitch  point.  Assume  no  backlash  and  the  width 
of  the  teeth  equal  to  the  spaces  on  the  pitch  circles.  Draw  the  ad- 
dendum circles  of  each  with  radius  equal  to  radius  of  pitch  circle 
plus  addendum.  Draw  root  circle  of  each  with  radius  equal  to  radius 
of  pitch  circle  minus  (addendum  of  other  gear  plus  clearance).  Space 
off  the  circular  pitch  on  either  side  of  P  on  each  pitch  circle.  This 
may  be  conveniently  done  by  drawing  a  line  tangent  to  the  pitch  circles 
at  P,  laying  off  the  circular  pitch  PC  and  PCi  on  this  line.  Set  the 
dividers  at  some  small  distance  such  that  when  spaced  on  the  pitch 
circles  the  length  of  arc  and  chord  will  be  nearly  the  same.  Start  at  C, 
step  back  on  CP  until  the  point  of  the  dividers  comes  nearly  to  P  (say 
at  K)  then  step  back  on  the  pitch  circles  the  same  number  of  spaces, 
getting  H  and  L.  Hi  and  LI  can  be  found  in  the  same  manner. 

Through  the  point  /,  where  the  curve  RT  cuts  the  pitch  circle  of  B, 
draw  the  radial  line  cutting  the  addendum  circle  at  V.  Make  arc  WX 


GEARS  AND  GEAR  TEETH 


103 


equal  to  arc  VR.  Cut  a  templet  or  find  a  place  on  a  French  curve  which 
fits  the  curve  RJT,  mark  it,  and  transfer  the  curve  to  pass  through  X 
and  P.  Make  PH%  equal  to  one  half  PHi,  turn  the  curve  over  and  draw 


FIG.  134 

curve  through  Hz  in  the  same  way  that  PX  was  drawn.    All  the  other 
curves  may  be  drawn  in  a  similar  way. 

126.  Clearing  Curve.     If  the  flanks  are  extended  until  they  join 
the  root  Line,  a  very  weak  tooth  will  often  result;  to  avoid  this,  a 


104 


ELEMENTS  OF  MECHANISM 


fillet  is  used  which  is  limited  by  the  arc  of  a  circle  connecting  the  root 
line  with  the  flank,  and  lying  outside  the  actual  path  of  the  end  of 
the  face  of  the  other  wheel.  This  actual  path  of  the  end  of  the  face 
is  called  the  true  clearing  curve. 


FIG.  135 

This  curve  is  the  epitrochoid  traced  by  the  outermost  corner  of  one 
tooth  on  the  plane  of  the  other  gear.  The  general  method  of  drawing 
such  a  curve  is  shown  in  Fig.  135.  The  tooth  M  is  to  work  in  the  space 
N.  From  e  lay  off  the  equal  arcs  eei,  e\e^}  e^,  etc.,  and  from  /  lay  off 


GEARS  AND  GEAR  TEETH 


105 


the  same  distance  //i,  //2,  etc.  From  /,  /i,  /2,  etc.,  draw  arcs  with  the 
radii  eR,  eiR,  e2R,  etc.,  respectively.  A  smooth  curve  internally  tangent 
to  all  these  curves  will  be  the  desired  epitrochoid  or  clearing  curve. 

127.  The  Involute  of  a  Circle.  The  form  of  the  curve  most  com- 
monly given  to  gear  teeth  is  that  known  as  the  involute  of  a  circle. 
Teeth  properly  constructed  with  this  curve  will  conform  to  the  law 
described  in  §  122  as  will  appear  in  the  following  paragraphs.  This 
curve  and  the  method  of  drawing  it  will,  therefore,  be  studied  before 
considering  the  method  of  applying  it  to  gear  teeth. 

In  Fig.  136  the  circle  represents  the  end  view  of  a  cylinder  around 
which  is  wrapped  an  inextensible  fine  thread,  fastened  to  the  cylinder 
at  A  and  having  a  pencil  in  a  loop  at  P.  If  now  the  pencil  is  swung 
out  so  as  to  unwind  the  thread 
from  the  cylinder,  keeping  it 
always  taut,  the  curve  which  the 
pencil  traces  on  a  piece  of  paper 
on  which  the  cylinder  rests  is 
known  as  an  involute  of  the 
circle  which  represents  the  end 
view  of  the  cylinder.  The  same 
result  is  obtained  by  considering 
the  tracing  point  to  be  carried 
by  a  line  rolling  on  a  circle.  All 
involutes  drawn  from  the  same 
circle  are  alike,  but  involutes 
drawn  from  circles  of  different 
diameters  are  different.  The 
greater  the  diameter  of  the  circle 
the  flatter  will  be  its  involute. 

In  constructing  the  involute 
of  a  circle  on  the  drawing  board 
it  is,  of  course,  impossible  actu- 
ally to  wrap  a  thread  around 
the  circle  and  draw  the  involute 
by  unwinding  the  thread.  Fig. 
137  shows  the  method  of  constructing  an  involute  on  the  drawing 
board.  Suppose  the  involute  is  to  be  drawn  starting  from  any  point  p 
on  the  circle  whose  center  is  C.  Set  the  dividers  at  any  convenient  short 
spacing;  a  distance  which  is  about  ^  the  circumference  of  the  circle 
will  give  good  results.  Place  one  of  the  points  of  the  dividers  at  p  and 
space  along  on  the  circumference  a  few  times,  getting  the  equidistant 
points  m,  n,  r,  s.  At  each  of  these  points  draw  radial  lines  and  con- 


FIG.  136 


FIG.  137 


106 


ELEMENTS  OF  MECHANISM 


struct  lines  perpendicular  to  these  radii  as  shown.  Each  of  these 
perpendiculars  will  then  be  tangent  to  the  circle  at  one  of  the  points. 
Taking  care  that  the  setting  of  the  dividers  remains  unchanged,  lay 
off  one  space  ml  on  the  tangent  at  m.  On  the  next  line,  which  is 
tangent  at  n,  lay  off  from  n  the  same  distance  twice,  getting  the  point 
2.  From  r  lay  off  the  distance  three  times,  getting  the  point  3;  and  so 
on  until  points  are  found  as  far  out  as  desired.  A  smooth  curve  drawn 
through  these  points  with  a  French  curve  will  be  a  very  close  approxi- 
mation to  the  true  involute,  —  close  enough  for  all  practical  purposes 
if  the  work  is  done  carefully. 

128.  Application  of  the  Involute  to  Gears.     In  Fig.  138  let  A  and  B 
be  the  centers  of  two  gears  whose  pitch  circles  are  tangent  at  P.    Through 


Circle 
'itch  Circle 


\  ^-^  Pitch  Circle 


Circle 


P  draw  a  line  XX  perpendicular  to  the  line  of  centers  A  B  and  another 
line  YY  making  an  angle  K  with  XX.  From  A  draw  a  line  Aa  perpen- 
dicular to  YY  and  from  B  draw  Bb  also  perpendicular  to  YY.  Then  Aa 
and  Bb  will  be  the  radii  of  circles  drawn  from  A  and  B  respectively,  tan- 
gent to  YY.  These  circles  are  called  base  circles.  The  triangle  AaP  is 

similar  to  the  triangle  BbP,  therefore,  -~  =  -gp .     That  is,  the  radii  of 

the  base  circles  are  in  the  same  ratio  as  the  radii  of  the  pitch  circles. 

Therefore,  since 

Angular  speed  of  A  _  BP 

Angular  speed  of  B      AP 


GEARS  AND  GEAR  TEETH  107 

it  follows  that  Angular  speed  of  A  =  Bb 

Angular  speed  of  B      Aa 

If  now  the  tooth  outlines  on  the  gear  A  are  made  involutes  of  the 
circle  whose  radius  is  A  a  and  those  on  B  involutes  of  the  circle  whose 
radius  is  Bb,  a  tooth  on  A  will  drive  a  tooth  on  B  in  such  a  way  that  at 
all  times  the  angular  speed  of  A  will  be  to  the  angular  speed  of  B  as 
Bb  is  to  A  a,  the  action  being  the  same  as  if  the  lines  ab  and  ai&i  were 
inextensible  cords  connecting  the  base  circles  and  the  involutes  were 
curves  traced  by  marking  points  on  the  cords.  The  same  ratio  of  speeds 
would  hold  if  B  were  the  driver.  The  teeth  would  always  be  in  contact  at 
a  point  on  the  line  aPb  or  at  a  point  on  aiPbi.  The  path  of  contact  in 
gears  having  involute  teeth  is,  therefore,  a  straight  line  and  the  angle  of 
obliquity  or  pressure  angle  is  constant.  That  is,  the  direction  of  the  force 
which  the  driving  tooth  exerts  on  the  driven  tooth  is  the  same  at  all  times. 

129.  To  Draw  a  Pair  of  Involute  Gears.  Suppose  that  it  is  required 
to  draw  a  pair  of  involute  gears  4-pitch,  16  teeth  in  the  driver  and  12 
teeth  in  the  driven  gear;  addendum  on  each  to  be  J  in.  and  dedendum 
A  in.;  pressure  angle  6  =  22|°. 

One-half  of  one  gear  and  4  teeth  on  the  other  will  be  drawn.  In 
Fig.  139  draw  a  center  line  and  on  this  line  choose  a  point  S  which  is 
to  be  the  center  of  the  driving  gear.  To  find  the  distance  between 
centers  and  thus  locate  the  center  of  the  other  gear,  first  find  the  pitch 
diameter  of  each.  Since  the  driver  has  16  teeth  and  is  4-pitch  (that  is, 
it  has  4  teeth  for  every  inch  of  pitch  diameter),  its  pitch  diameter  must 
be  16  -r-  4  or  4  in.  In  like  manner  the  diameter  of  the  other  gear  is 
12  -s-  4  or  3  in.  The  distance  between  centers  must  be  equal  to  the 
radius  of  the  driver  plus  the  radius  of  the  driven  gear  and  is,  therefore, 
2  +  1|  in.  or  3J  in.  Measure  off  the  distance  SSi  equal  to  3J  in.  and 
Si  is  the  center  of  the  driven  gear.  Next,  locate  the  pitch  point  P,  2  in. 
from  S  or  1J  in.  from  Si,  and  through  P  draw  arcs  of  circles  with  S  and 
Si  as  centers.  These  arcs  are  parts  of  the  pitch  circles  of  the  two  gears. 
Through  P  draw  the  line  XX  perpendicular  to  the  line  of  centers  and 
draw  the  line  YY  making  an  angle  of  22 1°  with  XX.  From  S  and  Si 
draw  lines  perpendicular  to  YY  meeting  it  at  a  and  b.  With  radii  Sa 
and  Sib  draw  the  base  circles.  Draw  the  addendum  circle  of  the  upper 
gear  with  S  as  a  center  and  radius  equal  to  the  radius  of  the  pitch  circle 
plus  the  addendum  distance.  This  will  be  2}  in.  In  similar  manner 
draw  the  addendum  circle  of  the  lower  gear  with  a  radius  If  in.  Draw 
the  root  circle  of  the  upper  gear  with  S  as  a  center  and  radius  2  —  -fa 
in.  (that  is  pitch  radius  —  dedendum).  Find  the  root  circle  of  the  lower 
gear  in  a  similar  way.  We  are  now  ready  to  construct  the  teeth. 


108 


ELEMENTS  OF  MECHANISM 


From  the  point  a  space  off  on  the  base  circle  the  arc  at  equal  in  length  to 
the  line  aP  and  from  t,  thus  found,  draw  the  involute  of  the  base  circle 
of  the  upper  gear  as  described  for  Fig.  137.  tPk  is  the  curve  thus  found. 
In  a  similar  manner  find  the  point  r  such  that  arc  br  is  equal  in  length 
to  the  line  bP  and  from  r  draw  the  involute  rPn  of  the  lower  base  circle. 

The  shape  of  the  tooth  curves  having  been  found  in  this  way,  the 
next  step  is  to  find  the  width  of  the  teeth  on  the  pitch  circles  and  draw 
in  the  remaining  curves.  Since  the  gears  are  4-pitch,  the  circular  pitch 


is  i  X  3.1416  =  0.7854  in.  and  if  the  width  of  the  tooth  is  one-half  the 
circular  pitch,  as  is  usually  the  case,  the  width  of  the  tooth  on  each  gear 
must  be  i  X  0.7854  or  0.39  in.  nearly.  Therefore,  lay  off  the  arc  PR 
equal  to  0.39  in.  and  through  R  draw  an  involute  which  is  a  duplicate 
of  the  curve  tPk  except  that  it  is  turned  in  the  reverse  direction.  Simi- 
larly make  PT  equal  0.39  and  draw  an  involute  through  T  which  is  a 
duplicate  of  the  curve  nPr.  These  curves  can  be  transferred  to  the  new 
positions  by  means  of  templets,  it  being  unnecessary  to  construct  the 
curve  more  than  once.  The  part  of  the  tooth  outlines  below  the  base 


GEARS  AND  GEAR  TEETH 


109 


circles  may  be  made  radial  lines  with  small  fillets  at  the  bottom  corners. 
One  tooth  on  each  gear  has  now  been  completed  and  other  teeth  may 
be  drawn  like  these  by  means  of  templets. 

If  the  larger  gear  is  the  driver  and  turns  in  the  direction  indicated  by 
the  arrow,  the  path  of  contact  is  the  line  MPN. 


w 


FIG.  140 

130.  Normal  Pitch.  The  normal  pitch  is  the  distance  from  one 
tooth  to  the  corresponding  side  of  the  next  tooth,  measured  on  the  com- 
mon normal  (CCi,  Fig.  140).  From  the  method  of  generating  the  curves 
this  distance  is  constant  and  is  equal  to  the  distance  between  the  corre- 
sponding sides  of  two  adjacent  teeth  measured  on  the  base  circle  (arc 
KKlt  Fig.  140). 


110  ELEMENTS  OF  MECHANISM 

The  definite  quantities  in  a  given  involute  gear  are  the  base  circle 
and  the  normal  pitch. 

131.  Relation  between  Normal  Pitch  and  Circular  Pitch.     Referring 
to  Fig.  140,  let  D  represent  the  diameter  of  the  pitch  circle  and  A,  the 
diameter  of  the  base  circle.    N  =  normal  pitch,  C  =  circular  pitch, 
T  the  number  of  teeth,  a  the  point  where  the  line  of  obliquity  (or  generat- 
ing line)  is  tangent  to  the  base  circle,  and  0  the  pressure  angle.     Draw 
Sa  and  produce  it  to  meet  XX  (the  tangent  to  the  pitch  circle  through 
P)  at  W.    Angle  aSP  =  angle  aPW  =  0,  and  the  triangles  aPW  and 
aSP,  are  similar. 

Therefore,  -~p  =  cos  0. 

From  the  definition  of  normal  pitch  (§  130) 

and  from  equation  (44) 

Therefore, .  ~-  =  j*  =  cos  0.  (50) 

That  is,  the  normal  pitch  is  equal  to  the  circular  pitch  multiplied  by  the 
cosine  of  the  pressure  angle. 

132.  Relation  between  Length  of  Path  of  Contact  and  Length  of 
Arc  of  Contact.     In  Fig.  141  the  teeth  shown  in  full  lines  are  in  con- 
tact at  the  beginning  of  the  path  of  contact  and  the  teeth  shown 
dotted  are  at  the  end  of  the  path  of  contact. 

The  angle  a  is  therefore  the  angle  of  action  on  the  driven  gear  and 
arc  NPM  is  the  arc  of  action.  The  corresponding  arc  of  the  base  circle 
isLK. 

As  previously  shown, 

Radius  base  circle 


Radius  pitch  circle 


=  cos 


.  arc  LK 

Therefore,  -  TTT>  =  cos  0. 

arc  NM 

But,  from  the  properties  of  the  involute,  considering  the  line  ab  as  the 
connecting  line  between  the  two  revolving  base  circles,  the  length  of 
the  line  TTi  (that  is,  the  path  of  contact)  is  equal  to  the  length  of  the 
arc  LK. 


Therefore,  arc  r    =  cos 


GEARS  AND  GEAR  TEETH 


111 


That  is,  the  path  of  contact  is  equal  to  the  arc  of  contact  multiplied  by  the 
cosine  of  the  pressure  angle. 

A  convenient  method  for  finding  this  relation  graphically  is  as  follows : 
Draw  XX  tangent  to  the  pitch  circles  at  P.    From  T  and  TI  draw  TR 


Driven 


FIG.  141 

and  TiS  perpendicular  to  TTi,  meeting  XX  at  R  and  S.  The  length 
RS  is  then  equal  to  the  length  of  the  arc  of  action,  RP  being  the  length 
of  the  arc  of  approach  and  PS  the  length  of  the  arc  of  recess. 


112 


ELEMENTS  OF  MECHANISM 


Conversely,  if  the  lengths  of  the  arcs  of  approach  and  recess  are 
known,  the  ends  of  the  paths  of  contact  may  be  found  by  laying  off  the 
arcs  along  XX  and  drawing  perpendiculars  to  ab. 

133.  Limits  of  Addendum  on  Involute  Gears.  Fig.  141  shows  one 
tooth  on  each  of  a  pair  of  4-pitch  gears  of  18  and  24  teeth  respectively. 


FIG.  142 


The  addendum  arcs  of  the  teeth,  shown  in  full  lines,  are  such  that  the 
addendum  distance  is  equal  to  f  the  module  (^  in.).  If  for  any  reason 
it  is  desired  to  redesign  these  gears  with  longer  teeth,  that  is,  with  larger 


GEARS  AND  GEAR  TEETH 


113 


addendum  circles,  it  will  be  necessary  to  know  how  long  the  teeth  can 
be  made  without  causing  trouble.  The  tooth  on  B  can  be  increased  in 
length  until  the  addendum  circle  passes  through  the  point  a,  where  the 
line  of  obliquity  Y  Y  is  tangent  to  the  base  circle  of  A .  If  the  tooth  is 
made  longer  than  this  limit,  interference  will  result  unless  some  special 
form  of  curve  is  constructed  in  place  of  the  involute  for  the  outer  end 
of  the  tooth. 

At  the  right  of  Fig.  142  are  shown  the  same  pair  of  teeth  with  the 
addendum  of  gear  B  lengthened  so  that  the  addendum  circle  is  outside 
of  point  a.  It  will  be  noticed  that  the  extended  face  of  the  tooth 


B 
j  22  Teeth 


FIG.  142a 


of  B  cuts  into  the  radial  extension  of  the  flank  of  the  tooth  on  A 
and  also  cuts  slightly  into  that  part  of  the  tooth  outside  the  base 
circle. 

To  show  that  interference  of  this  sort  occurs  refer  to  Fig.  142a  where 
a  22-tooth  gear  B  is  shown  working  with  a  10-tooth  pinion  A.  P  is  the 
pitch  point,  a  the  point  where  the  base  circle  of  the  pinion  is  tangent 
to  the  line  of  obliquity.  If  the  addendum  of  the  gear  B  is  carried  beyond 
a  (say,  to  k)  the  conjugate  to  the  line  ok  is  the  involute  na  of  the  pinion 
base  circle  and  lies  in  the  space  between  the  pinion  teeth.  To  see  that 
this  is  true  assume  that  the  line  of  obliquity  is  a  cord  carrying  a  point 
which  traces  the  involutes  (see  §  128)  and,  instead  of  wrapping  around 
the  base  circle  of  the  pinion,  is  tangent  to  it  at  a  and  extends  beyond 


114  ELEMENTS  OF  MECHANISM 

carrying  a  marking  point  at  WL  As  the  gear  B  turns  to  the  left  the  cord, 
by  its  contact  at  a,  turns  the  base  circle  of  the  pinion  right-handed, 
the  tracing  point  HI  traces  the  outline  kiOy  (same  as  ka)  on  the  plane  of 
the  gear  B  and  the  curve  n^a^  (same  as  no)  on  the  plane  of  the  pinion  A. 

Next,  to  show  that  the  curve  kiOy  will  cut  into  curve  a\t\  (the  regular 
pinion  tooth)  as  the  tracing  point  approaches  a,  assume  the  tracing 
point  to  be  at  n*.  The  curve  k20w  (same  as  ka)  will  be  tangent  to  o^nw 
(same  as  an)  at  712.  Now  ofa  is  tangent  to  cwiw  at  a*.  Therefore,  since 
k&w  and  ofa  are  tangent  to  the  same  curve  a^nw  at  different  points,  their 
directions  must  be  such  that  they  tend  to  intersect.  In  the  figure  the 
interference  is  evident.  When  the  tracing  point  reaches  a  the  inter- 
ference ceases. 

The  above  consideration  shows  that  while  the  common  normal  will 
trace  conjugate  curves  beyond  a,  it  is  impossible  to  get  proper  action 
for  that  portion  of  the  tooth  outline  kar  outside  of  a,  first  because  the 
conjugate  to  ka  lies  inside  the  tooth  of  which  kar  is  the  outline,  and  sec- 
ond because  the  part  of  the  tooth  of  which  ka  is  the  outline  cuts  into 
the  tooth  outline  at. 

Referring  again  to  Fig.  142,  the  tooth  on  A  might  be  lengthened  until 
the  addendum  circle  passed  through  the  point  of  tangency  b  except  for 
the  fact  that  there  is  another  limit  to  the  addendum  which  sometimes 
has  to  be  considered.  The  maximum  addendum  here  is  limited  by  the 
intersection  of  the  two  sides  of  the  tooth  giving  a  pointed  tooth.  It  is 
evident  that  no  further  increase  in  addendum  is  here  possible. 

The  following  illustration  will  help  to  make  the  above  statements 
clear. 

In  Fig.  143  let  it  be  required  to  determine  if  the  arc  of  recess  can 
be  equal  to  f  of  the  circular  pitch.  Lay  off  from  a  on  the  tangent  the 
distance  ab  =  f  of  the  circular  pitch.  Draw  be  perpendicular  to  the 
line  of  obliquity;  c  will  be  the  end  of  the  path  of  contact  for  the  given 
arc  of  recess.  If  the  point  c  came  beyond  d,  the  tangent  point  of  the 
line  of  obliquity  and  the  base  circle,  the  action  would  be  impossible  since 
no  contact  can  occur  beyond  d.  But  if,  as  in  Fig.  143,  the  point  c  comes 
between  a  and  d,  it  is  necessary  to  determine  if  the  face  of  .the  tooth 
on  A  can  reach  to  c.  Lay  off  on  the  pitch  circle  A  the  arc  ae  =  ab  =  f 
of  the  pitch;  the  face  of  the  tooth  on  A  will  then  pass  through  c  and  e. 
Draw  the  line  co  from  c  to  the  center  of  A,  and  note  the  point  /  where  it 
cuts  the  pitch  circle  A.  If  ef  is  less  than  one-half  the  thickness  of  the 
tooth,  the  action  can  go  as  far  as  c  and  the  teeth  will  not  be  pointed. 
In  the  figure,  assuming  tooth  and  space  equal,  the  thickness  of  the  tooth 
would  be  eg,  and  ef  is  less  than  \  eg;  therefore  the  action  is  possible,  as 
is  shown  by  the  two  teeth  drawn  in  contact  at  c. 


GEARS  AND  GEAR  TEETH 


115 


FIG.  143 


116 


ELEMENTS  OF  MECHANISM 


134.  Involute  Pinion  and  Rack.  Fig.  144  shows  a  pinion  driving 
a  rack.  The  path  of  contact  cannot  begin  before  the  point  a,  but  the 
recess  is  not  limited  excepting  by  the  addendum  of  the  pinion,  since 
the  base  line  of  the  rack  is  tangent  to  the  line  of  obliquity  at  infinity. 
For  the  same  reason  it  will  be  evident  that  the  sides  of  the  teeth  of  the 
rack  will  be  straight  lines  perpendicular  to  the  line  of  obliquity.  In  the 
figure  the  addendum  on  the  rack  is  made  as  much  as  the  pinion  will 
allow,  that  is,  so  that  the  path  of  contact  will  begin  at  a.  The  adden- 
dum of  the  pinion  will  give  the  end  of  the  path  of  contact  at  b. 


FIG.  144 

In  Fig.  145,  the  diagram  for  a  pinion  and  a  rack,  let  it  be  required 
to  determine  if  the  path  of  contact  can  begin  at  a  and  go  as  far  as  6; 
to  be  solved  without  using  the  tooth  curves.  For  the  contact  to  begin 
at  a  the  face  of  the  rack  must  reach  to  a.  Draw  the  line  ac  perpendicu- 
lar to  the  line  of  obliquity,  giving  cd  as  the  arc  of  approach;  draw  ae 
parallel  to  the  line  of  centers,  and  if  ce  is  less  than  one-half  the  thick- 
ness of  the  rack  tooth,  the  approaching  action  is  possible  without 
pointed  teeth.  Similarly  for  the  recess,  draw  the  line  bf  perpendicular 
to  the  line  of  obliquity,  giving  df  equal  to  the  arc  of  recess;  make  the 
arc  dg  on  the  pinion's  pitch  circle  equal  to  df,  then  the  face  of  the  pinion's 
tooth  will  pass  through  b  and  g',  draw  the  line  bh  to  the  center  of  the 
pinion,  and  note  the  point  h  where  it  crosses  the  pinion's  pitch  circle. 
If  gh  is  less  than  one-half  the  thickness  of  the  tooth,  the  recess  is  pos- 
sible without  pointed  teeth, 


GEARS  AND  GEAR  TEETH 


117 


135.  Involute  Pinion  and  Annular  Wheel.  Fig.  146  shows  an  invo- 
lute pinion  driving  an  annular  wheel.  This  case  is  very  similar  to  a 
pinion  and  rack.  The  addendum  of  the  annular  is  limited  by  the  tan- 
gent point  a  of  the  pinion's  base  circle  and  the  line  of  obliquity,  while 
the  addendum  of  the  pinion  is  unlimited  except  by  the  teeth  becoming 
pointed.  The  base  circle  of  the  annular  lies  inside  the  annular,  so  that 
its  point  of  tangency  with  the  line  of  obliquity  is  at  b.  If  we  take  some 
point  on  the  line  of  obliquity,  as  c,  and  roll  the  tooth  curves  as  they 
would  appear  in  contact  at  that  point,  the  teeth  of  the  annular  will  be 


FIG.  145 

found  to  be  concave,  and  the  addendum  of  the  annular  would  seem  to  be 
limited  by  the  base  circle  of  the  annular  where  the  curves  end.  But  if 
these  two  teeth  are  moved  back  until  they  are  in  contact  at  a,  it  will  be 
evident  that  the  annular's  tooth  curve  cannot  be  extended  beyond  a 
without  interfering  with  the  pinion  teeth  as  in  the  case  of  the  gear  in 
Fig.  142a.  Therefore  the  addendum  of  the  annular  is  limited  by  the 
point  of  tangency  of  the  base  circle  of  the  pinion  and  the  line  of  obliquity. 
If  the  ratio  of  the  number  of  teeth  in  the  pinion  to  the  number  of 
teeth  in  the  annular  exceeds  a  certain  limit,  interference  will  occur 


118 


ELEMENTS  OF  MECHANISM 


between  the  teeth  after  they  have  ceased  contact  along  the  path  of  con- 
tact.    This  is  illustrated  in  Fig.  146a,  where  an  18-tooth  pinion  is  shown 


OQ 


FIG.  146 


with  a  24-tooth  annular  of  14J°  obliquity.    Their  teeth  are  shown  inter- 
fering at  K. 

The  limiting  size  of  the  pinion  at  which  this  interference  begins  to  be 
evident  is  a  function  of  the  angle  of  obliquity. 


GEARS  AND  GEAR  TEETH 


119 


136.  Possibility  of  Separating  two  Involute  Wheels.  Interchange- 
able Gears.  One  of  the  most  important  features  of  involute  gearing 
is  the  fact  that  two  such  wheels  may  be  separatedj  within  limits,  without 
destroying  •  the  accuracy  of  the  angular  speed  ratio.  In  this  way  the 
backlash  may  be  adjusted,  since  the  original  pitch  circles  need  not  be 
in  contact.  To  show  that  this  is  so,  the  gears  shown  in  Fig.  147  may 
be  redrawn  using  the  same  pitch  circles  and  base  circles,  but  separating 
them  slightly,  keeping  the  teeth  in  contact,  as  has  been  done  in  Fig. 
148.  Connect  the  base  circles  by  the  tangent  be.  If  now  the  line  be 
carries  a  marking-point,  it  will  evidently  trace  the  involutes  of  the  two 
base  circles,  as  de  and  he,  and  these  curves  must  be  the  same  as  the  tooth 


FIG.  146a 

curves  in  Fig.  147.  In  Fig.  148  these  curves  de  and  he  will  give  an 
angular  speed  ratio  to  the  base  circles  inversely  as  their  radii,  but  the 
radii  of  these  base  circles  are  directly  as  the  radii  of  the  original  pitch 
circles  (Fig.  147) ;  hence  in  Fig.  148  the  tooth  curves  de  and  he  would 
give  an  angular  speed  ratio  to  the  two  wheels  inversely  as  the  radii  of 
the  original  pitch  circles,  although  these  circles  do  not  touch.  The  path 
of  contact  is  now  from  k  to  e,  which  is  considerably  shorter  than  in 
Fig.  147;  it  is,  however,  slightly  more  than  the  normal  pitch,  so  that  the 
action  is  still  sufficient.  The  limit  of  the  separation  will  be  when  the 
path  of  contact  is  just  equal  to  the  normal  pitch.  The  pressure  angle 
is  6am,  which  is  greater  than  in  Fig.  147.  The  backlash  has  also 
increased. 


120 


ELEMENTS  OF  MECHANISM 


The  wheels  have  new  pitch  circles  in  contact  at  a,  and  a  new  angle 
of  obliquity  or  pressure  angle,  also  a  greater  circular  pitch  with  a  certain 


o/q 


amount  of  backlash;  and  if  these  latter  data  had  been  chosen  at  first 
the  result  would  have  been  exactly  the  same  wheels  as  in  Fig.  147,  slightly 
separated.  It  will  be  seen  that  the  radii  of  the  new  pitch  circles  are  to 


GEARS  AND  GEAR  TEETH 


121 


each  other  as  the  radii  of  the  respective  base  circles,  and  consequently 
as  the  respective  original  pitch  circles.     It  will  also  be  seen  that  the 


FIG.  148 

line  of  obliquity,  which  is  the  common  normal  to  the  tooth  curves,  passes 
through  the  new  pitch  point  a  so  that  the  fundamental  law  of  gearing 
is  still  fulfilled. 


122  ELEMENTS  OF  MECHANISM 

By  the  application  of  the  preceding  principles  two  or  more  wheels 
of  different  numbers  of  teeth,  turning  about  one  axis,  can  be  made  to 
gear  correctly  with  one  wheel  or  one  rack;  or  two  or  more  parallel  racks 
with-  different  obliquities  of  action  may  be  made  to  gear  correctly  with 
one  wheel,  the  normal  pitches  in  each  case  being  the  same.  Thus  dif- 
ferential movements  may  be  obtained  which  are  not  possible  with  teeth 
of  any  other  form. 

In  this  same  connection,  attention  may  be  called  to  the  fact  that  in 
a  set  of  involute  gears  which  are  to  be  interchangeable  the  normal  pitch 
must  be  the  same  in  all. 

137.  Standard  Proportions.     There  is  no  one  standard  governing 
the  relations  between  pitch,  addendum,  clearance,  etc.     Two  methods 
of  proportioning  the  teeth  may  be  mentioned  which,  for  convenience, 
will  be  referred  to  as  the  Brown  &  Sharpe  Standard  and  the  American 
Society  of  Mechanical  Engineers  Standard.     The  Brown  &  Sharpe 
standard  represents  the  proportions  ordinarily  used  by  the  Brown  & 
Sharpe  Manufacturing  Company.     Their  practice  is  to  modify  the  form 
of  the  tooth  curves  slightly  at  the  end  to  avoid  interference,  or  as  expe- 
rience has  shown  them  to  be  desirable.     The  A.S.M.E.  standard  is  that 
proposed  in  a  majority  report  of  a  committee  appointed  to  recommend 
a  standard  which  would  be  desirable  for  general  adoption. 

The  following  tables  give  the  proportion  for  the  two  standards. 

TABLE  I.  — BROWNE  &  SHARPE  STANDARD  FOR 
INVOLUTE  GEARS 

Angle  of  obliquity  (pressure  angle) 14£° 

Addendum Equal  to  module 

Clearance Approximately  |  module 

Dedendum  or  root Approximately  1|  module 

TABLE  II.  — A.S.M.E.  STANDARD  (PROPOSED)  FOR 
INVOLUTE  GEARS 

Angle  of  obliquity  (pressure  angle) 22^° 

Addendum I  module 

Clearance I  module 

Dedendum  or  root Equal  to  module 

Another  standard,  differing  slightly  from  the  A.S.M.E.  standard, 
gives  the  form  of  tooth  known  as  the  stub  tooth. 

138.  Cycloidal  Gears.     Formerly,  gear  teeth  were  constructed  on 
the  cycloidal  system.     The  faces  of  the  teeth  were  epicycloids  generated 
on  the  pitch  circles  and  the  flanks  hypocycloids  generated  inside  the 
pitch  circles.     The  involute  system  has  replaced  the  cycloidal  almost 


GEARS  AND  GEAR  TEETH 


123 


entirely  for  general  purposes,  although  cycloidal  teeth  are  still  used  in 
some  special  cases. 

In  Fig.  149  let  01  and  o2  be  the  centers  of  the  two  wheels  A  and  B, 
their  pitch  circles  being  in  contact  at  the  point  a.  Let  the  smaller  circles 
C  and  D,  with  centers  at  pi  and  pz,  be  placed  so  that  they  are  tangent 
to  the  pitch  circles  at  a.  Assume  the  centers  of  these  four  circles  to  be 
fixed  and  that  they  turn  in  rolling  contact;  then  if  the  point  a  on  the 


circle  A  moves  to  a\,  02,  as,  the  same  point  on  B  will  move  to  61,  62,  &s, 
and  on  C  to  ci,  c%,  c3.  Now  if  the  point  a  on  the  circle  C  carries  a  marking- 
point,  in  its  motion  to  Ci  it  will  have  traced  from  the  circle  A  the  hypo- 
cycloid  aid,  and  at  the  same  time  from  the  circle  B  the  epicycloid  bid. 
This  can  be  seen  to  be  true  if  the  circles  A  and  B  are  now  fixed;  and  if 
C  rolls  in  A,  the  point  Ci  will  roll  to  ai,  tracing  the  hypocycloid  c\a\'t 
while  if  C  rolls  on  B,  ci  will  trace  the  epicycloid  CI&L  These  two  curves 


124        .  ELEMENTS  OF  MECHANISM 

in  contact  at  c\  fulfil  the  fundamental  law  for  tooth  curves,  that  the 
normal  to  the  two  curves  at  the  point  Ci  must  pass  through  a.  Simi- 
larly, if  the  original  motion  of  the  circles  had  been  to  02,  62,  €2,  the  same 
curves  would  be  generated,  only  they  would  be  longer  and  in  contact 
at  02.  If  the  hypocycloid  c^Os,  is  taken  for  the  flank  of  a  tooth  on  A,  and 
the  epicycloid  czbz  for  the  face  of  a  tooth  on  B,  and  if  c^az  drives  Czbz 
toward  a,  it  is  evident  that  these  two  curves  by  their  sliding  action,  as 
they  approach  the  line  of  centers,  will  give  the  same  type  of  motion  to 
the  circles  as  the  circles  had  in  generating  the  curves,  which  was  pure 
rolling  contact.  Therefore  the  two  cycloidal  curves  rolled  simultane- 
ously by  the  describing  circle  C  will  cause  by  their  sliding  contact  the 
same  angular  speed  ratio  of  A  and  B  as  would  be  obtained  by  A  and  B 
moving  with  pure  rolling  contact. 

If  now  the  circles  A,  B,  and  D  are  rolled  in  the  opposite  direction  to 
that  taken  for  A,  B,  and  C,  and  if  the  point  a  moves  to  a4,  64,  and  di  on 
the  respective  circles,  the  point  a  on  D  while  moving  to  di  will  trace 
from  A  the  epicycloid  a^di,  and  from  B  the  hypocycloid  Wi.  The  curve 
didi  may  be  the  face  of  a  tooth  on  A,  and  Mi  the  flank  of  a  tooth 
on  B,  the  normal  d\a  to  the  two  curves  in  contact  at  di  passing  through  a. 
The  flank  and  face  for  the  teeth  on  A  and  B,  respectively,  which  were 
previously  found,  have  been  added  to  the  face  and  flank  just  found, 
giving  the  complete  outlines,  in  contact  at  d\. 

If  now  the  wheel  B  is  turned  L.H.,  the  tooth  shown  on  it  will  drive 
the  tooth  on  A,  giving  a  constant  angular  speed  ratio  between  A  and  B 
until  the  face  of  the  tooth  on  B  has  come  to  the  end  of  its  action  with 
the  flank  which  it  is  driving,  at  about  the  point  c^. 

The  following  facts  will  be  evident  from  the  foregoing  discussion:  in 
the  cycloidal  system  of  gearing,  the  flank  and  face  which  are  to  act  upon 
each  other  must  be  generated  by  the  same  describing  circle,  but  the  describ- 
ing circles  for  the  face  and  flank  of  the  teeth  of  one  wheel  need  not  be 
alike.  The  path  of  contact  is  always  on  the  describing  circles;  in  Fig.  149 
it  is  along  the  line  diaci. 

139.  Interchangeable  Wheels.  A  set  of  wheels  any  two  of  which 
will  gear  together  are  called  interchangeable  wheels.  For  these  the 
same  describing  circle  must  be  used  in  generating  all  the  faces  and 
flanks.  The  size  of  the  describing  circle  depends  on  the  properties  of 
the  hypocycloid,  which  curve  forms  the  flanks  of  the  teeth  (excepting 
in  an  annular  wheel).  If  the  diameter  of  the  describing  circle  is  half 
that  of  the  pitch  circle,  the  flanks  will  be  radial  (Fig.  150,  A),  which 
gives  a  comparatively  weak  tooth  at  the  root.  If  the  describing  circle 
is  made  smaller,  the  hypocycloid  curves  away  from  the  radius  (Fig. 
150,  B)  and  will  give  a  strong  form  of  tooth;  but  if  the  describing  circle 


GEARS  AND  GEAR  TEETH  125 

is  larger,  the  hypocycloid  will  curve  the  other  way,  passing  inside  the 
radial  lines  (Fig.  150,  C)  and  giving  a  still  weaker  form  of  tooth,  and 
a  form  of  tooth  which  may  be  impossible  to  shape  with  a  milling- 
cutter. 

From  the  above  the  practical  conclusion  would  appear  to  be  that  the 
diameter  of  the  describing  circle  should  not  be  more  than  one-half  that 
of  the  pitch  circle  of  the  smallest  wheel  of  the  set.  It  will  be  found, 
however,  that  when  the  diameter  of  the  describing  circle  is  taken  five- 
eighths  the  diameter  of  the  pitch  circle,  the  curvature  of  the  flanks  will 
not  be  so  great,  with  the  ordinary  proportions  of  height  to  thickness  of 
teeth,  but  that  the  spaces  are  narrowest  at  the  bottom;  this  being  the 
case,  the  teeth  can  be  shaped  by  a  milling-cutter 

In  one  set  of  wheels  in  common  use  the  diameter  of  the  describing 
circle  is  taken  such  that  it  will  give  radial  flanks  on  a  15-tooth  pinion, 


FIG.  150 

or  five-eighths  that  of  a  12-tooth  pinion,  the  smallest  wheel  of  the  set. 
This  describing  circle  has  been  used  with  excellent  results. 

As  an  example,  given  an  interchangeable  set  of  cycloidal  gears,  2-P., 
radial  flanks  on  a  15-tooth  pinion;  a  gear  having  24  teeth  is  to  drive 
one  having  30  teeth.  The  diameter  of  a  2-P.,  15-tooth  pinion  would  be 
1\  inches;  to  give  radial  flanks  on  this  pinion  the  diameter  of  the  de- 
scribing circle  would  be  3f  inches.  This  is  the  diameter  of  the  describ- 
ing circle  for  all  the  faces  and  flanks  for  any  gear  of  the  set.  The  24-tooth 
gear  will  have  a  diameter  of  12",  and  the  30-tooth  gear  will  have  15" 
diameter.  This  will  give  the  diagram  in  Fig.  151  ready  for  the  rolling 
of  the  tooth  outlines. 

140.  To  draw  the  Teeth  for  a  Pair  of  Cycloidal  Wheels,  and  to 
determine  the  Path  of  Contact.  In  Fig.  152  given  the  pitch  circles 
A  and  B  and  the  describing  circles  C  and  D,  C  to  roll  the  faces  for  B 


126 


ELEMENTS  OF  MECHANISM 


and  the  flanks  for  A,  while  D  is  to  roll  the  faces  for  A  and  the  flanks 
for  B.  These  curves  may  be  rolled  at  any  convenient  place.  In  the 
figure,  the  wheel  A  is  to  be  the  driver  and  is  to  turn  as  shown.  Choose 
any  point,  as  6,  on  A  and  a  point  a  on  B  at  a  distance  from  the  pitch 
point  af  =  bf.  The  epicycloid  and  hypocycloid  rolled  from  a  and  b 
respectively,  and  shown  in  contact  at  62,  would  be  suitable  for  the  faces 
of  the  teeth  on  B  and  the  flanks  of  the  teeth  on  A  respectively,  and  could 
be  in  action  during  approach.  The  curves  may  be  rolled  as  indicated 
by  the  light  lines.  The  method  used  to  roll  these  curves  is  shown  in 
Fig.  153,  where  the  circle  C  is  tracing  a  hypocycloid  on  A  from  the 

point  o.  Assume  the  circle  C  to  start 
tangent  to  A  at  o  and  to  roll  as  shown, 
drawing  it  in  as  many  positions  as  may 
be  desired  to  obtain  a  smooth  curve, 
and  these  positions  do  not  need  to  be 
equidistant;  thus  in  the  figure  the 
center  of  C  is  at  b,  c,  and  d  for  the  three 
positions  used.  Since  the  circle  C  rolls 
on  A,  the  distance  measured  on  A  from 
o  to  a  tangent  point  of  C  and  A  iis 
equal  to  the  distance  measured  on  C 
from  that  tangent  point  to  the  hypo- 
cycloid.  The  method  of  spacing  off 
these  equal  arcs  for  the  successive 
positions  is  indicated  in  Fig.  153. 

Returning  to  Fig.  152,  the  circle  D 
is  to  roll  the  faces  for  the  teeth  on  A 
and  the  flanks  for  the  teeth  on  B. 
These  curves  may  also  be  rolled  from 
any  convenient  points,  as  c  and  d  equi- 
distant from  /.  The  face  thus  found  from  A  may  be  traced  and  then 
transferred  to  the  flank  already  found  for  the  teeth  on  A  at  the  point  b, 
giving  the  curve  62&c',  the  entire  acting  side  of  a  tooth  on  A.  Similarly 
by  transferring  the  flank  dd3  to  the  point  a  we  have  b%ad',  the  shape  of 
the  teeth  for  the  wheel  B.  It  will  be  seen  that  the  face  on  A  could  have 
been  rolled  from  b  as  well  as  from  c,  so  that  the  entire  tooth  curve  could 
be  rolled  from  b,  and  similarly  the  other  tooth  curve  could  have  been 
rolled  from  the  point  a.  After  finding  the  tooth  curves,  and  knowing 
the  addendum,  clearance,  and  backlash,  the  teeth  may  be  drawn.  In 
Fig.  152  the  teeth  are  drawn  without  backlash,  and  in  contact  on  their 
acting  surfaces  at  h  and  k.  The  path  of  contact  is  efg  on  the  describing 
circles  and  is  limited  by  the  addendum  circles. 


FIG.  151 


GEARS  AND  GEAR  TEETH 


127 


141.  Limits  of  the  Path  of  Contact.  Possibility  of  Any  Desired 
Action.  If,  in  Fig.  152,  the  teeth  of  either  wheel  are  made  longer,  the 
path  of  contact  and  arc  of  action  are  increased;  the  extreme  limit  of 
the  path  of  contact  would  therefore  be  when  the  teeth  become  pointed. 

It  is  often  desirable  to  find  whether  a  desired  arc  of  action  in  approach 
or  in  recess  may  be  obtained  before  rolling  the  tooth  curves.  Given 
the  pinion  A  driving  the  rack  B  as  shown  in  Fig.  154;  to  determine  if 

\ 


FIG.  152 

an  arc  of  approach  equal  to  ab  is  possible.  The  path  of  contact  must 
then  begin  at  c,  where  the  arc  ac  is  equal  to  ab.  The  face  of  the  rack's 
tooth  must  be  long  enough  to  reach  from  b  to  c,  and  this  depends  on  the 
thickness  of  the  tooth  measured  on  the  pitch  line,  since  the  non-acting 
side  of  the  tooth  must  not  cause  the  tooth  to  become  pointed  before 
the  point  c  is  reached.  To  see  if  this  is  possible  without  the  tooth  curves, 
draw  a  line  from  c  parallel  to  the  line  of  centers  (in  general  this  line  is 
drawn  to  the  center  of  the  wheel;  the  rack's  center  being  at  infinity  gives 


128 


ELEMENTS  OF  MECHANISM 


the  line  parallel  to  the  line  of  centers),  and  note  the  point  d  where  this 
line  crosses  the  pitch  line  of  the  rack.  If  bd  were  just  equal  to  one-half 
the  thickness  of  the  tooth,  the  tooth  would  be  pointed  at  c,  and  the 
desired  arc  of  approach  would  be  just  possible;  if  bd  were  kss  than  one- 
half  the  thickness  of  the  tooth,  the  tooth  would  not  become  pointed 
until  some  point  beyond  c  was  reached,  so  that  the  action  would  be  pos- 
sible and  the  teeth  not  pointed,  as  shown  by  the  figure. 

If  it  is  desired  to  have  the  arc  of  recess  equal  to  the  arc  of,  then  the 
path  of  contact  must  go  to  g,  and  the  face  of  the  pinion  must  remain  in 
contact  with  the  flank  of  the  rack  until  that  point  is  reached,  or  the 
face  must  be  long  enough  to  reach  from  /  to  g.  Drawing  a  line  from  g 
to  the  center  of  the  pinion  A,  we  find  that  the  distance  fh  is  greater 


FIG.  153 

than  one-half  of  fk,  which  is  taken  as  the  thickness  of  the  tooth;  there- 
fore the  desired  arc  of  recess  is  not  possible  even  with  pointed  teeth. 

The  minimum  arc  of  action,  as  in  the  case  of  involute  gears,  is  the 
circular  pitch. 

142.  Annular  Wheels.  Fig.  155  shows  a  pinion  A  driving  an  annu- 
lar wheel  B,  the  describing  circle  C  generating  the  flanks  of  A  and  the 
faces  of  B,  which  in  an  annular  wheel  lie  inside  the  pitch  circle,  while  D 
generates  the  faces  of  A  and  the  flanks  of  B.  The  describing  circle  C 
is  called  the  interior  describing  circle,  and  D  is  called  the  exterior  describ- 
ing circle.  The  method  of  rolling  the  tooth  curves,  and  the  action  of 


GEARS  AND  GEAR  TEETH 


129 


the  teeth,  are  the  same  as  in  the  case  of  two  external  wheels,  the  path  of 
contact  being  in  this  case  efg  when  the  pinion  turns  R.H.  If  these 
wheels  were  of  an  interchangeable  set,  the  describing  circles  would  be 
alike  and  found  as  explained  in  §  139,  and  the  annular  would  then  gear 


FIG.  154 

with  any  wheel  of  the  set  excepting  for  a  limitation  which  is  discussed 
in  the  following  paragraph. 

143.  Limitation  in  the  Use  of  an  Annular  Wheel  of  the  Cycloidal 
System.  Referring  to  Fig.  155,  it  will  be  evident  that,  if  the  pinion 
drives,  the  faces  of  the  pinion  and  annular  will  tend  to  be  rather  near  each 
other  during  recess  (during  approach  also  on  the  non-acting  side  of  the 


130  ELEMENTS  OF  MECHANISM 

teeth).  The  usual  conditions  are  such  that  the  faces  do  not  touch;  but 
the  conditions  may  be  such  that  the  faces  will  touch  each  other  without 
interference,  for  a  certain  arc  of  recess;  or,  finally,  the  conditions  may 
be  such  that  the  faces  would  interfere,  which  would  make  the  action  of 
the  wheels  impossible. 

To  determine  whether  a  given  case  is  possible  it  is  necessary  to  refer 
to  the  double  generation  of  the  epicycloid  and  of  the  hypocycloid. 
The  acting  face  of  the  pinion,  Fig.  155,  is  rolled  by  the  exterior  describ- 
ing circle  D,  while  the  acting  face  of  the  annular  is  generated  by  the 
anterior  describing  circle  C.  Two  such  faces  are  shown  in  Fig.  156  as 
they  would  appear  if  rolled  from  the  points  g  and  h,  equidistant  from 
the  pitch  point  k.  The  acting  face  of  A  is  an  epicycloid,  and  is  made 
by  rolling  the  circle  D  to  the  right  on  A.  It  is  also  true  that  a  circle 
whose  diameter  is  equal  to  the  sum  of  the  diameters  of  A  and  D  would 
roll  the  same  epicycloid  if  rolled  in  the  same  direction.  This  circle  is 
E,  Fig.  156,  and  is  called  the  intermediate  describing  circle  of  the  pinion. 
The  acting  face  of  the  annular  is  a  hypocycloid  rolled  by  the  interior 
describing  circle  C  rolling  to  the  left  inside  of  B.  It  is  also  true  that 
the  same  hypocycloid  would  be  rolled  by  a  circle  whose  diameter  is 
equal  to  the  difference  between  the  diameters  of  B  and  C,  provided  it  is 
rolled  in  the  opposite  direction.  This  circle  is  F,  Fig.  156,  and  is  called 
the  intermediate  describing  circle  of  the  annular. 

If  now  the  four  circles  A,  B,  E,  and  F  turn  in  rolling  contact,  through 
arcs  each  equal  to  kg,  the  point  k  will  be  found  at  g,  h,  m,  and  n  on  the 
respective  circles,  the  point  k  on  E  having  rolled  the  epicycloid  gm, 
while  k  on.  F  rolls  the  hypocycloid  hn. 

To  determine  whether  these  faces  do  or  do  not  touch  or  conflict, 
assume  that  the  given  conditions  gave  the  circles  E  and  F  coincident 
as  in  Fig.  157  where 

diam.  A  +  diam.  D  =  diam.  E  =  diam.  B  -  diam.  C  =  diam.  F. 

Here  if  the  three  circles  A,  B,  and  (EF)  turn  in  rolling  contact,  the 
point  k  moving  to  g  on  A  will  move  to  h  on  B  and  to  (mn)  on  the  com- 
mon intermediate  circle.  This  means  that  the  common  intermediate 
circle  could  simultaneously  generate  the  two  faces;  therefore  the  two 
faces  are  in  perfect  contact  on  the  intermediate  circle.  This  contact 
will  continue  until  the  addendum  circle  of  one  of  the  wheels  crosses 
the  intermediate  circle,  the  addendum  circle  crossing  first  necessarily 
limiting  the  path  of  contact. 

The  above  may  be  stated  as  follows:  //  the  intermediate  describing 
circles  of  the  pinion  and  annular  coincide,  the  faces  will  be  in  contact  in 
recess,  if  the  pinion  drives,  in  addition  to  the  regular  path  of  contact. 


GEARS  AND  GEAR  TEETH 


131 


FIG.  155 


132 


ELEMENTS  OF  MECHANISM 


GEARS  AND  GEAR  TEETH 


133 


If  in  Fig.  157  the  exterior  describing  circle,  for  example,  should  be 
made  smaller,  as  in  Fig.  158,  then  the  intermediate  of  the  pinion  would 
be  smaller  than  that  of  the  annular;  but  if  the  exterior  describing  circle 
is  smaller,  the  face  gm  will  have  a  greater  curvature  and  will  evidently 
curve  away  from  the  face  hn,  so  that  no  contact  between  the  faces  can 
occur,  as  is  shown  in  Fig.  158.  Here  no  additional  path  of  contact 


FIG.  157 

occurs,  and  it  is  evident,  if  the  arcs  kg,  km,  kn,  and  kh  are  equal  as  they 
must  be,  if  the  circles  move  in  rolling  contact,  that  the  smaller  D  be- 
comes (and  consequently  E)  the  greater  will  be  the  space  between  the 
faces. 

This  may  be  stated  as  follows:  //  the  intermediate  describing  circle  of 
the  pinion  is  smaller  than  that  of  the  annular,  the  faces  do  not  touch,  and 
the  action  is  in  all  respects  similar  to  the  cases  of  external  wheels. 


134 


ELEMENTS  OF  MECHANISM 


In  Fig.  159  the  exterior  describing  circle  D  is  made  larger  than  it  is 
in  Fig.  157,  so  that  the  intermediate  E  of  the  pinion  is  larger  than  F, 
that  of  the  annular.  Making  the  circle  D  larger  would  give  the  face  of 

1 


FIG.  159 


the  pinion  less  curvature,  which  would  cause  tne  curve  gm  to  cross  the 
curve  hn,  giving  an  impossible  case.  Therefore,  if  the  intermediate  of 
the  pinion  is  greater  than  that  of  the  annular,  the  action  is  impossible. 


GEARS  AND  GEAR  TEETH 


135 


144.  Low-numbered  Pinions,  Cycloidal  System.  The  obliquity  of 
action  in  cycloidal  gears  is  constantly  varying;  it  diminishes  during  the 
approach,  becoming  zero  at  the  pitch  point,  and  then  increases  during 
the  recess.  For  wheels  doing  heavy  work  it  has  been  found  by  experi- 
ence that  the  maximum  obliquity  should  not  in  general  exceed  30°,  giving 
a  mean  of  15°.  When  more  than  one  pair  of  teeth  are  in  contact,  a  high 
maximum  is  less  objectionable. 

As  the  number  of  teeth  in  a  wheel  decreases,  they  necessarily  become 
longer  to  secure  the  proper  path  of  contact,  and  both  the  obliquity  of 


FIG.  160 

action  and  the  sliding  increase.  From  the  preceding  considerations  the 
practical  rule  is  deduced  that,  for  millwork  and  general  machinery,  no 
pinion  of  less  than  twelve  teeth  should  be  used  if  it  is  possible  to  avoid  it. 

It  often  becomes  necessary,  however,  to  use  wheels  having  less  than 
twelve  teeth,  in  light-running  mechanism,  such  as  clockwork.  In  such 
cases  a  greater  obliquity  may  be  admissible,  and  for  light  work  the  flank- 
describing  circle  may  be  made  large. 

Let  it  be  required  to  determine  the  possibility  of  using  two  equal 
pinions,  having  six  teeth,  with  radial  flanks,  the  arcs  of  approach  and 


136 


ELEMENTS  OF  MECHANISM 


recess  each  equal  to  one-half  the  pitch,  and  to  find  the  maximum  angle 
of  obliquity.  Fig.  160  is  the  diagram  for  two  such  gears.  The  path  of 
contact  is  to  begin  at  a,  the  arcs  ab,  cb,  and  db  each  being  equal  to  one- 
half  the  pitch;  then  the  face  of  the  pinion  B  must  be  long  enough  to  be 
in  contact  with  the  flank  of  A  at  a.  Drawing  the  line  aef  from  a  to  the 
center  of  B,  the  distance  de  will  be  found  to  be  less  than  one-half  the 
thickness  of  the  tooth.  Therefore  the  approach  is  possible.  Since  the 
pinions  are  alike,  the  recess  is  also  possible.  The  maximum  angle  of 


FIG.  161 

obliquity  in  approach  is  the  angle  8,  and  this  may  be  found  in  degrees 
as  follows.  The  arc  be  on  the  pitch  circle  A  subtends  an  angle  bgc  equal 
to  one-half  the  pitch  angle,  the  arc  of  approach  being  equal  to  one-half 
the  pitch;  in  this  case  the  angle  bgc  is  30°.  The  arc  ab  on  the  describ- 
ing circle  C  is  equal  to  be  and  therefore  subtends  an  angle  bha,  which  is 
to  the  angle  bgc  inversely  as  the  radii  of  the  respective  circles.  In  this 
case  these  radii  are  as  2  to  1,  making  the  angle  bha  equal  to  60°.  (It  is 
important  to  notice  that  the  line  gc  does  not  pass  through  the  point  a 
excepting  in  the  single  case  of  a  radial  flank  gear.)  The  angle  B  between 
the  tangent  and  the  chord  ab  will  always  be  one-half  the  angle  ahb  sub- 
tended by  the  arc  ab.  This  gives  the  angle  of  obliquity  30°.  Therefore 
we  find  that  two  pinions  with  six  teeth  and  radial  flanks  will  work  with 


GEARS  AND  GEAR  TEETH 


137 


arcs  of  approach  and  recess  each  equal  to  one-half  the  pitch  and  with 
the  maximum  angle  of  obliquity  of  30°.  By  allowing  a  greater  angle  of 
obliquity  the  teeth  may  be  made  a  little  longer  and  so  give  an  arc  of 
action  greater  than  the  pitch,  which  should  be  the  case  in  practice. 

Two  pinions  with  five  teeth  each  will  work  with  describing  circles 
having  diameters  three-fifths  the  diameter  of  the  pitch  circles,  and  arcs 
of  approach  and  recess  each  equal  to  one-half  the  pitch,  as  shown  by 


FIG.  162 

Fig.  161,  the  path  of  contact  beginning  at  a,  the  arcs  ab,  cb,  and  db  each 
being  equal  to  one-half  the  pitch.  The  action  is  possible,  since  de  is 
less  than  one-half  the  thickness  of  the  tooth.  The  maximum  angle  of 
obliquity  is  30°,  the  angle  bgc  being  36°  and  bha  being  f  of  36°,  or  60°. 
Two  pinions  with  four  teeth  each  will  just  barely  work  with  describ- 
ing circles  having  diameters  five-eighths  the  diameter  of  the  pitch  circle, 
and  with  no  backlash,  the  arcs  of  approach  and  recess  each  being  one- 
half  the  pitch.  Fig.  162  shows  the  diagram  for  this  case,  and  the  teeth 


138  ELEMENTS  OF  MECHANISM 

are  apparently  pointed,  which  would  be  the  case  if  de  were  just  one-half 
the  thickness  of  the  tooth.  To  determine  the  possibility  of  the  action 
the  angle  dfe  may  be  calculated.  It  should  not  be  greater  than  22J° 
to  allow  the  desired  arc  of  approach.  It  will  be  found  to  be  22°  27'  19", 
so  that  the  action  is  just  possible.  The  maximum  angle  of  obliquity 
6  will  be  found  to  be  36°. 

A  pinion  with  four  teeth  will  work  with  a  pinion  having  four  teeth 
or  any  higher  number,  if  the  arc  of  action  is  not  required  to  be  greater 
than  the  pitch,  the  maximum  angle  of  obliquity  not  exceeding  36°. 

The  requirements  may  be  very  different  from  the  above  in  every 
respect;  an  arc  of  action  greater  than  the  pitch  would  usually  be  re- 
quired; it  might  be  desired  to  have  the  arc  of  recess  greater  than  the 
arc  of  approach;  it  might  not  be  admissible  to  have  so  great  an  angle 
of  obliquity  or  to  have  the  teeth  cut  under  so  far  as  a  describing  circle 
five-eighths  the  pitch  circle  would  require.  The  results  would  of  course 
vary  with  the  conditions  imposed. 

145.  Stepped  Wheels.     If  a  pair  of  spur  wheels  are  cut  transversely 
into  a  number  of  plates,  and  each  plate  is  rotated  through  an  angle, 
equal  to  the  pitch  angle  divided  by  the  number  of  plates,  ahead  of  the 
adjacent  plate,  as  shown  in  Fig.  163,  the  result  will  be  a  pair  of  stepped 
wheels.     This  device  has  the  effect  of  increasing  the  number  of  teeth 

.      ........      V/,,,,A __ — ,     without  diminishing  their  strength;   and 

f  WM^A?  \      \  •  • 

> — bWlr^  i^Jf^f™  t^iM^ — \    the  number  of  contact  points  is  also  in- 
creased.  The  upper  figure  shows  a  section 
on  the  pitch  line  A  A.    The  action  for 
each  pair  of  plates  is  the  same  as  that  for 
spur  wheels  having  the  same  outlines.    In 
practice  there  is  a  limit  to  the  reduction 
in  the  thickness  of  the  plates,  depending 
on  the  material  of  the  teeth  and  the  pres- 
sure to  be  transmitted,  since  too  thin  plates  would  abrade.     The  number 
of  divisions  is  not  often  taken  more  than  two  or  three,  and  the  teeth 
are  thus  quite  broad.    These  wheels  give  a  very  smooth  and  quiet  action. 

146.  Twisted  Spur  Gears.     If,  instead  of  cutting  the  gear  into  a 
few  plates,  as  shown  in  Fig.  163,  the  number  of  sections  is  infinite,  the 
result  is  a  helical  gear  such  as  that  shown  in  Fig.  116. 

The  twisting  being  uniform,  the  elements  of  the  teeth  become  helices, 
all  having  the  same  lead,  see  §§  161  and  162.  The  line  of  contact 
between  two  teeth  will  have  a  helical  form,  but  will  not  be  a  true  helix; 
the  projection  of  this  helix  on  a  plane  perpendicular  to  the  axis  will  be 
the  ordinary  path  of  contact.  It  can  easily  be  seen  that  the  common 
normal  at  any  point  of  contact  can  in  no  case  lie  in  the  plane  of  rota- 


• 


GEARS   AND  GEAR  TEETH 


139 


tion,  but  will  make  an  angle  with  it.  The  line  of  action  then  can  in 
general  have  three  components:  1°  A  component  producing  rotation, 
perpendicular  to  the  plane  of  the  axes;  2°  A  component  of  side  pressure, 
parallel  to  the  line  of  centers;  3°  A  component  of  end  pressure  parallel 
to  the  axes.  The  end  pressure  may  be  neutralized  as  explained  in 
§147. 

147.  Herring-bone  Gears.     A  gear  like  that  shown  in  Fig.  117, 
known  as  a  herring-bone  gear,  is  equivalent  to  two  helical  gears,  one 
having  a  right-hand  helix  and  the  other  a  left-hand  helix.     The  use  rf 
a  pair  of  gears  of  this  type  eliminates  the  end 

thrust  on  the  shaft  referred  to  in  the  preceding 
paragraph. 

148.  Sliding  Friction  Eliminated.     In  Fig. 
164,  which  represents  a  transverse  section  of  a 
pair  of  twisted  wheels,  suppose  the  original  tooth 
outlines  to  have  been  those  shown  dotted.  Then 
cut  away  the  faces  as  shown  by  full  lines  having 
the  new  faces  tangent  to  the  old  ones  at  the 
pitch  point  c;  proper  contact  is  lost  except  that 

at  c  for  the  section  shown,  but  by  twisting  the  wheels  this  contact  can 
be  made  to  travel  along  the  common  element  of  the  pitch  cylinders 
through  c  from  one  side  of  the  wheel  to  the  other.  A  simple  con- 
struction to  use  in  this  case  is  to  make  the  flanks  of  the  wheels  radial 
and  the  faces  semicircles  tangent  to  the  flanks.  The  action  here  is 
purely  rolling  and  is  very  smooth  and  noiseless; 
but  for  heavy  work  it  is  best  to  use  the  common 
forms  of  teeth  with  sliding  action,  so  that  the 
pressure  may  be  distributed  over  a  line  instead 
of  acting  at  a  point. 

149.  Pin  Gearing.  In  this  form  of  gearing 
the  teeth  of  one  wheel  consist  of  cylindrical 
pins,  and  those  of  the  other  of  surfaces  parallel 
to  cycloidal  surfaces,  from  which  they  are 
derived. 

In  Fig.  165  let  01  and  02  be  the  centers  of 
the  pitch  circles  whose  circumferences  are  di- 
vided into  equal  parts,  as  ce  and  eg.  Now  if 
we  suppose  the  wheels  to  turn  on  their  axes, 
and  to  be  in  rolling  contact  at  c,  the  point  e  of  the  wheeloi  will  trace 
the  epicycloid  gp  on  the  plane  of  the  wheel  02,  and  merely  a  point  e 
upon  the  plane  of  o\.  Let  cf  be  a  curve  similar  to  ge  and  imagine  a  pin 
of  no  sensible  diameter  —  a  rigid  material  line  —  to  be  fixed  at  c  in  the 


FIG.  165 


140 


ELEMENTS  OF  MECHANISM 


upper  wheel.  Then,  if  the  lower  one  turn  to  the  right,  it  will  drive  the 
pin  before  it  with  a  constant  velocity  ratio,  the  action  ending  at  e  if 
the  driving  curve  be  terminated  at  /as  shown. 

If  the  pins  be  made  of  a  sensible  diameter,  the  outlines  of  the  teeth 
upon  the  other  wheel  are  curves  parallel  to  the  original  epicycloids,  as 
shown  in  Fig.  166.  The  diameter  of  the  pins  is  usually  made  about 
equal  to  the  thickness  of  the  tooth,  the  radius  being,  therefore,  about  J 
the  pitch  arc.  This  rule  is,  however,  not 
imperative,  as  the  pins  are  often  made  con- 
siderably smaller. 

Clearance  for  the  pin  is  provided  by  forming 
the  root  of  the  tooth  with  a  semicircle  of  a 
radius  equal  to  that  of  the  pin,  the  center  being 
inside  of  the  pitch  circle  an  amount  equal  to 
the  clearance  required. 

The  pins  are  ordinarily  supported  at  each 
end,  two  discs  being  fixed  upon  the  shaft  for 
the  purpose,  thus  making  what  is  called  a 
lantern  wheel  or  pinion. 

In  wheel  work  of  this  kind  the  action  is 
almost  wholly  confined  to  one  side  of  the  line 
of  centers.  In  the  elementary  form  (Fig.  165) 
the  action  is  wholly  on  one  side,  and  receding,  since  it  cannot  begin  until 
the  pin  reaches  c  (if  o2  drives),  and  ceases  at  e;  if  o\  is  considered  the 
driver,  action  begins  at  e,  ends  at  c,  and  is  wholly  approaching.  As 
approaching  action  is  injurious,  pin  gearing  is  not  adapted  for  use  where 
the  same  wheel  has  both  to  drive  and  to  follow;  the  pins  are  therefore 
always  given  to  the  follower,  and  the  teeth  to  the  driver. 

When  the  pin  has  a  sensible  diameter,  the  tooth  is  shortened  and 
its  thickness  is  decreased;  the  line  of  action  is  also  shortened  at  e,  Fig. 
166,  and,  instead  of  beginning  at  c,  will  begin  at  a  point  where  the 
normal  to  the  original  tooth  curve,  through  the 
center  of  the  pin,  first  comes  in  contact  with 
the  derived  curve  mf.  This  normal's  end  will 
not  fall  at  c,  but  at  a  point  on  the  arc  ce  beyond, 
on  account  of  a  property  of  the  curve  parallel  to 
an  epicycloid.  The  parallel  to  the  epicycloid  is 
shown  in  Fig.  167,  cp  being  the  given  epicy- 
cloid. The  curve  may  be  found  by  drawing 
a  series  of  arcs  ss  with  a  radius  equal  to  the  normal  distance  between 
the  curves,  and  with  the  centers  on  cp.  The  parallel  curve  first 
passes  below  the  pitch  curve  cm  and  then  rises,  after  forming  a  cusp, 


FIG.  166 


FIG.  167 


GEARS  AND   GEAR  TEETH 


141 


and  cuts  away  the  first  part  drawn:  this  is  more  clearly  shown 
somewhat  exaggerated  at  mno.  Hence  the  part  which  would  act  on 
the  pin  when  its  center  is  at  c  is  cut  away,  and,  for  the  same  epicycloid, 
the  greater  the  diameter  of  the  pin  the  more  this  cutting  away.  In 
Fig.  166  the  pin  e  is  just  quitting  contact  with  the  tooth  at  i  while  c  is 
at  the  pitch  point,  and,  according  to  the  above  property  of  the  parallel 
to  the  epicycloid,  is  not  yet  in  contact  with  the  tooth  m.  Strictly  speak- 
ing, then,  the  case  shown  is  not  a  desirable  one,  as  the  tooth  should 
not  cease  contact  at  i  until  m  begins  its  action.  The  above  error  is 
practically  so  small  that  it  has  been  disregarded,  especially  for  rough 
work. 

The  following  method  may  be  used  in  determining  a  limiting  case 
in  pin  gearing: 

If  the  pitch  arc  =  eg  is  assumed  (Fig.  168),  the  greatest  possible 
height  of  tooth  is  determined  by  the  intersection  of  the  front  and  back 
of  the  tooth  at  p;  and  if  this  height 
is  taken,  action  will  begin  at  c  and 
end  at  h,  the  point  in  the  upper 
pitch  circle  through  which  p  passes. 
Now  if  p  falls  upon  the  pitch  circle 
ceh,  we  should  have  a  limiting  case 
for  a  pin  of  no  sensible  diameter. 
If  the  pin  has  a  sensible  diameter 
and  the  pitch  arc  eg  =  ce  is  assigned, 
bisect  eg  with  the  line  o^p  and  draw 
ce  intersecting  o^p  in  fc;  assume  a 
radius  for  the  pin  less  than  ek  and 
draw  the  derived  curve  to  cut  Ozp 
in  j,  which  will  be  the  point  of  the 
tooth.  Through  j  draw  a  normal 
to  the  epicycloid,  cutting  it  at  s; 
through  s  describe  an  arc  about  02 
cutting  the  upper  pitch  circle  at  t, 
the  position  of  the  center  of  the  pin  at 
the  end  of  its  action.  Draw  the  outline  mf  of  the  next  working  tooth, 
find  the  point  m  at  the  cusp  of  the  curve  parallel  to  the  epicycloid,  and 
draw  the  normal  ran;  m  is  the  lowest  possible  working  point  of  the 
tooth.  Through  n  describe  an  arc  about  o2  cutting  the  original  path 
of  contact  in  r,  which  is  the  point  that  n  must  reach  before  the  tooth  will 
be  in  contact  with  the  pin,  or  is  the  point  that  n  must  reach  before  the 
common  normal  to  the  pin  and  tooth  curve  passes  through  the  pitch 
point. 


FIG.  168 


142  ELEMENTS  OF  MECHANISM 

Now  action  begins  when  the  axis  of  the  pin  is  at  r  and  ends  at  t;  if 
rt  =  ce,  we  have  an  exact  limiting  case  and  the  assumed  radius  of  the 
pin  is  a  maximum;  if  rt  <  ce,  the  radius  is  too  great;  but  if  rt  >  ce, 
the  case  is  practical  To  get  the  exact  limit  a  process  of  trial  and  error 
should  be  resorted  to.  When  the  pin  is  a  point  the  methods  used  in 
cycloidal  gearing  may  be  applied;  the  correction  for  a  pin  of  sensible 
diameter  can  then  be  made  by  applying  the  method  of  Fig.  168. 

150.  Pin  Gearing:  Wheel  and  Rack.     As  the  pins  are  always  given 
to  the  follower,  two  cases  arise. 

1°  Rack  drives,  giving  the  pin-wheel  and  rack,  Fig.  169.  Here  the 
original  tooth  is  bounded  by  cycloids  generated  by  the  pitch  circle  of 
the  wheel. 

2°  Wheel  drives,  giving  the  pin-rack  and  wheel.  Here  (Fig.  170)  the 
original  tooth  outline  is  the  involute  of  the  wheel's  pitch  circle. 

151.  Inside  Pin  Gearing.     Here  also  there  are  two  cases. 

1°  Pinion  drives  (Fig.  171).  The  original  tooth  outlines  will  be  inter- 
nal epicycloids  generated  by  rolling  the  pitch  circle  of  the  annular 
wheel  on  the  pinion's  pitch  circle. 

2°  Annular  wheel  drives  (Fig.  172).  Here  the  original  tooth  outline  is 
the  hypocycloid  traced  by  rolling  the  pinion's  pitch  circle  in  the  wheel's 
circle. 

Path  of  Contact.  In  the  elementary  form  of  tooth  (Fig.  165)  the  path 
of  contact  is  on  the  circumference  of  the  pitch  circle  of  the  follower  o\, 
as  ce.  When  a  pin  is  used  its  center  always  lies  in  this  circumference, 
and  its  point  of  contact  may  be  found  by  laying  off  a  distance  ei  equal 
to  the  radius  of  the  pin  (Fig.  173)  on  the  common  normal.  Drawing  a 
number  of  these  common  normals,  all  of  which  must  pass  through  the 
pitch  point  c,  and  laying  off  the  radius  of  the  pin  ei  on  each,  we  have 
the  path  of  contact  ci  known  as  the  limaQon. 

152.  Double-point  Gearing.     This  form  of  gearing,  shown  in  Fig. 
174,  gives  very  smooth  action  where  not  much  force  is  to  be  trans- 
mitted.    The  pitch  circles  are  here  taken  as  the  describing  circles;  the 
face  eg  of  the  pinion  01  is  generated  by  rolling  the  pitch  circle  o2  on  that 
of  0i,  and  the  face  cf  is  generated  by  rolling  the  pitch  circle  o\  on  o2. 
If  0i  is  considered  the  driver,  action  begins  at  d,  the  point  c  of  01  sliding 
down  the  face  cf  while  c  travels  from  d  to  c.     In  the  receding  action  the 
point  c  of  the  tooth  of  02  is  acted  on  by  the  face  eg  while  c  moves  from 
c  to  e.     The  spaces  must  be  so  made  as  to  clear  the  teeth.     This  com- 
bination reduces  friction  to  a  minimum  and  gives  the  obliquity  of  action 
less  than  in  any  case  except  pin  gearing,  but  the  teeth  are  much  under- 
cut and  weakened  by  the  clearing  curves,  and  if  much  force  is  to  be 
transmitted  the  line  of  contact  will  soon  be  worn  away. 


GEARS  AND  GEAR  TEETH 


143 


FIG.  169 


FIG.  170 


FIG.  171 


FIG.  172 


FIG.  173 


144 


ELEMENTS  OF  MECHANISM 


153.  Bevel  Gears.     A  pair  of  bevel  gears  bears  the  same  relation 
to  a  pair  of  rolling  cones  that  a  pair  of  spur  gears  bears  to  rolling  cylin- 
ders. 

Fig.  175  shows  two  bevel  gears  meshing  together,  the  contact  in  this 
case  being  internal.  Here,  as  in  the  case  of  spur  gears,  the  angular 
speeds  are  inversely  proportional  to  the  number  of  teeth. 

The  pitch  circle  of  a  bevel  gear  is  the  base  of  the  cone  which  the  gear 
replaces. 

154.  To  Draw  the  Blanks  for  a  Pair  of  Bevel  Gears.     A  convenient 
way  to  gain  an  understanding  of  the  principle  of  bevel  gear  design  will 

be  to  study  the  method  of 
drawing  the  blanks  from 
which  a  pair  of  bevel  gears  is 
to  be  cut.  Let  it  be  assumed 
that  a  6-pitch,  12-tooth  gear 
is  to  mesh  with  an  18-tooth 
gear,  the  axes  to  intersect  at 
90°.  Start  with  the  point  0, 
Fig.  176,  as  the  point  of  inter- 
section of  the  two  axes.  Draw 
OS  and  OSi  making  the  re- 
quired angle  (in  this  case  90°). 
These  are  the  center  lines  of 
the  shafts.  Assume  that  the 
12-tooth  gear  is  to  be  on  Si. 
Call  this  gear  A  and  the  18- 
tooth  gear  B.  Since  A  has 
12  teeth  and  is  6-pitch,  its 
pitch  diameter,  that  is,  the 
diameter  of  the  base  of  its 
pitch  cone,  is  12  -f-  6  or  2  in. 
In  like  manner  the  pitch 
diameter  of  B  is  18  -f-  6  or 
3  in.  From  0  measure  along 
OSi  a  distance  OM  equal  to  the  pitch  radius  of  B  (1J  in.)  and,  through 
M,  thus  found,  draw  a  line  perpendicular  to  OSi.  In  like  manner  make 
ON  equal  to  the  pitch  radius  of  A  and  draw  a  line  through  N  perpen- 
dicular to  OS.  These  lines  intersect  at  K.  Make  MR  equal  to  MK  and 
NT  equal  to  NK.  From  R,  K,  and  T  draw  lines  to  0.  Then  the  tri- 
angle ORK  is  the  projection  of  the  " pitch  cone"  of  the  gear  A  and  OTK 
that  of  the  " pitch  cone"  of  B.  It  will  be  noticed  that  the  above  con- 
struction is  the  equivalent  of  that  for  rolling  cones.  Next,  draw  through 


FIG.  175 


GEARS  AND  GEAR  TEETH 


145 


K  a  line  perpendicular  to  OK  meeting  OS!  at  P  and  OS  at  H  (not  shown). 
Draw  a  line  from  H  through  T  and  from  P  through  R.  The  cone  rep- 
resented by  the  triangle  THK  is  called  the  normal  cone  of  the  gear  B 
and  that  represented  by  the  triangle  RPK  is  called  the  normal  cone  of  A. 
These  cones  will  be  explained  more  in  detail  later. 

From  R  lay  off  Ra  equal  to  the  addendum  that  is  to  be  used  on  gear 
A  (this  is  determined  by  the  same  considerations  that  would  be  used 

t 


FIG.  176 

for  the  addendum  on  a  spur  gear).  Along  RO  lay  off  Rr  equal  to  the 
desired  length  of  gear  face  (§113).  Through  r  draw  a  line  parallel  to 
PR.  From  a  draw  a  line  to  0  meeting  this  parallel  at  ai.  Through  a 
draw  a  line  parallel  to  RK  meeting  PKH  at  02.  From  ai  draw  a  line 
parallel  to  RK  meeting  a  line  drawn  from  02  to  0  at  a3.  Lay  off  along 
RP  the  distance  Rd  equal  to  the  dedendum  and  draw  from  d  toward  0 
meeting  air  at  d\.  Find  d2  and  d3  in  the  same  way  that  #2  and  a3  were 


146  ELEMENTS  OF   MECHANISM 


found.  The  figure  addidi  represents  the  tooth.  The  dimensions  of  the 
hub  and  the  position  of  lines  FG  and  FiGi  and  of  the  corresponding 
lines  on  the  other  gear  may  be  made  anything  that  is  desirable. 

It  will  appear  from  this  construction  that  bevel  gears  must  be  laid 
out  in  pairs. 

155.  Teeth  for  Bevel  Gears.  The  gear  blanks,  as  laid  out  in  §  154, 
are  of  the  form  ordinarily  used,  and  the  information  there  given  is  per- 
haps all  that  a  person  making  use  of  bevel  gears  would  need.  In  order 
to  understand  the  principles  underlying  the  action  of  the  gears  it  may 
be  desirable  to  notice  the  relation  between  bevel  gear  teeth  and  spur 
gear  teeth. 

In  the  discussion  on  the  teeth  of  spur  wheels,  the  motions  were  con- 
sidered as  taking  place  in  the  plane  of  the  paper,  and  lines  instead  of 
surfaces  have  been  dealt  with.  But  the  pitch  and  describing  curves, 
and  also  the  tooth  outlines,  are  but  traces  of  surfaces  acting  in  straight- 
line  contact,  and  having  their  elements  perpendicular  to  the  plane  of 
the  paper.  In  bevel  gearing  the  pitch  surfaces  are  cones,  and  the  teeth 
are  in  contact  along  straight  lines,  but  these  lines  are  perpendicular  to 
a  spherical  surface,  and  all  of  them  pass  through  the  center  of  the 
sphere,  which  is  at  the  point  of  intersection  of  the  axes  of  the  two  pitch 
cones. 

In  Fig.  177,  0  is  the  center  of  the  sphere,  AOC  and  BOC  are  the  pitch 
cones.  If  the  teeth  are  involute,  cones  such  as  MON  and  KOL  are  the 

base  cones,  and  the  teeth  may  be  thought 
of  as  being  generated  by  a  plane  rolling 
on  each  of  the  base  cones,  the  ends  of  the 
teeth  lying  on  the  surface  of  the  sphere, 
and  the  tooth  outlines  being  the  curves 
traced  on  this  surface  by  the  plane  which 
generates  the  teeth. 

156.    Drawing  the   Teeth   on   Bevel 
Gears.  Tredgold's  Approximation.  Since 
^       \U-P  narrow  zones  of   the    sphere,  Fig.   177, 
K  near  the  circles  BC  and  AC,  will  nearly 

JPIG   177  coincide  with   cones  whose  elements  are 

tangent  at  B,  C  and  M,  the  conical  sur- 
faces may  be  substituted  for  the  spherical  ones  without  serious  error, 
and  as  the  tooth  outlines  are  always  comparatively  short  they  may 
be  supposed  to  lie  on  the  cones.  These  cones  BPC  and  ARC  are  called 
the  normal  cones  and  correspond  to  the  cones  RPK  and  THK  of  Fig. 
176.  Fig.  178  shows  the  method  of  drawing  the  tooth  outlines.  It  will 
be  noticed  that  the  developments  of  portions  of  the  normal  cones  are 


GEARS  AND  GEAR  TEETH 


147 


treated  as  if  they  were  pitch  circles  of  spur  gears  and  the  teeth  are 
drawn  on  the  development  exactly  as  if  they  were  teeth  of  spur  gears, 
and  are  then  transferred  to  the  other  views  by  ordinary  principles  of 
projection. 

157.  Crown  Gears.  When  the  angle  at  the  apex  of  the  cone  of  one 
of  a  pair  of  bevel  gears  is  180°  the  pitch  cone  becomes  a  flat  disk  and 
the  normal  cone  becomes  a  cylinder.  Such  a  gear  is  analagous  to  a 
rack  bent  in  the  form  of  a  circle.  The  teeth  taper  inward,  elements  of 


FIG.  178 

the  teeth  converging  toward  the  center  of  the  disk.  Another  bevel  of 
any  number  of  teeth  may  be  designed  to  run  with  a  crown  gear  but  the 
angle  between  the  axes  will  depend  upon  the  ratio  of  the  teeth.  Fig. 
179  shows  such  a  pair  of  gears. 

158.  Twisted  Bevel  Gears.     The  teeth  of  bevel  gears  may  be 
twisted  in  the  same  manner  as  the  teeth  of  spur  gears  (see  §  146).     Fig. 
122  shows  a  pair  of  twisted  bevels  such  as  used  for  the  drive  to  the 
differential  of  an  automobile.     The  gears  from  which  the  drawing  was 
made  were  from  a  Pierce  Arrow  touring  car. 

159.  Skew  Bevels.     Fig.  123  shows  a  pair  of  skew  bevel  gears  used 
in  cotton  machinery.     Here  the  shafts  are  at  right  angles,  non-inter- 


148 


ELEMENTS  OF  MECHANISM 


secting,  but  passing  so  near  each  other  that  ordinary  helical  gears  can- 
not be  used. 

The  pitch  surfaces  of  these  gears  are  hyperboloids  of  revolution,  and 
the  teeth  are  in  contact  along  straight  lines.  The  angular  speeds  are 
inversely  as  the  pitch  diameters. 

160.  Screw  Gearing.     This  class  of  gearing  is  used  to  connect  non- 
parallel  and  non-intersecting  shafts  and  includes  the  two  types  known 
as  worm  and  wheel  and  helical  gears.     The  latter  when  used  for  this 
purpose  are  often,  but  inaccurately,  called  spiral  gears.     In  the  helical 

gears  and  the  elementary  forms 
of  worm  and  wheel  the  teeth 
have  point  contact.  The  speed 
ratio  is  not  necessarily  in  the 
inverse  ratio  of  the  diameters. 
The  action  of  gears  of  this  class 
is  similar  to  the  action  of  a 
screw  and  nut  which  will  be 
considered  in  a  later  chapter. 
This  is  particularly  evident  in 
the  case  of  the  worm  and 
wheel.  The  distinction  be- 
tween the  worm  and  wheel  and 
the  helical  gears,  however,  is 
not  a  very  clear  one,  being 
largely  a  matter  of  speed  ratio 
and  manner  of  forming  the 
teeth.  Both  may  properly  be 
considered  here  as  helical 
gears  and  the  following  dis- 
FlG>  179  cussion  will  apply  to  both 

The  worm  and  wheel  will  be  considered  in  Chapter  VIII  as  a  screw 

and  nut. 

161.  The  Helix;    Its  Construction  and  Properties.    A  helix  is  a 
curve  wound  around  the  outside  of  a  cylinder  or  cone  advancing  uni- 
formly along  the  axis  as  it  winds  around.     The  nature  of  the  curve 
and  the  method  of  drawing  it  may  be  understood  from  a  study  of 
Fig.  180. 

The  angle  of  a  helix  is  the  angle  which  a  straight  line  tangent  to  the 
helix  at  any  point  makes  with  an  element  of  the  cylinder.  This  angle 
is  the  same  for  all  points  on  a  cylindrical  helix.  Two  helices  are  said 
to  be  normal  to  each  other  when  their  tangents  drawn  at  the  point 
where  the  helices  intersect  are  perpendicular  to  each  other. 


GEARS  AND  GEAR  TEETH 


149 


When  two  parallel  lines  wind  around  a  cylinder  forming  parallel 
helices,  as  in  Fig.  181,  the  result  would  be  called  a  double  helix;  three 
lines  would  give  a  triple  helix,  and  so  on.  If  the  helix  slopes  as  in  Fig. 


FIG.  180 

180  it  is  called  a  right-hand  helix;  if  it  slopes  in  the  reverse  direction 
it  is  called  a  left-hand  helix. 

162.  Lead.     Axial  Pitch.     The  distance  L,  Figs.  180  and  181,  by 
which  a  helix  advances  along  the  axis  of  the  cylinder  for  one  turn 
around  is  called  the  lead.     The  distance  A,  measured  parallel  to  the 
axis,  from  one  point  on  a  helix  to  the  corresponding  point  on  the 
next  turn  of  a  single  helix,  or  to  the  corresponding  point  on  the  next 
helix  in  the  case  of  a  multiple  helix,  is  called 

the  axial  pitch.  In  the  case  of  a  single  helix 
this  is  equal  to  the  lead;  in  the  case  of  a 
double  helix  the  axial  pitch  is  equal  to  one- 
half  the  lead;  in  a  triple  helix,  one-third 
of  the  lead,  and  so  on. 

163.  Normal   Pitch.     The   distance  P, 
between  a  point  on  a  helix   to  the  corre- 
sponding point  on  the  next  turn  of  a  single 
helix   or  the  corresponding  point   on    the 

next  helix  in  the  case  of  a  multiple  helix,  measured  along  the  normal  to 
the  helix,  is  called  the  normal  pitch. 

164.  Helical  Gears  are  gears  whose  teeth  wind  partially  around  the 
pitch  cylinders.     A  pair  of  such  gears  may  be  used  to  connect  parallel 
shafts,  as  shown  in  Fig.  116,  or  non-parallel  shafts,  which  is  the  case 
now  under  discussion.     The  method  of  forming  the  teeth  and  the 
action  of  the  teeth  differ  in  the  two  cases.*     The  definitions  given 

*  The  twisted  gears  (Fig.  116)  have  line  contact  between  teeth  while  the  helical 
gears  in  general  have  point  contact,  or  multiple  point  contact. 


FIG.  181 


150 


ELEMENTS  OF  MECHANISM 


above  apply  to  the  teeth  of  helical  gears.  In  order  that  two  helical 
gears  may  work  together  they  must  have  the  same  normal  pitch  and 
the  angle  between  the  shafts  must  be  such  that  the  tangent  to  the  pitch 
helices  of  the  two  gears  coincide  at  the  pitch  point.  From  this  it  follows 

Drivers  SJ&1 


that  the  sum  of  the  angles  of  their  helices  must  be  equal  to  the  angle 
between  the  shafts,  or  the  supplement  of  this  angle. 


k 


FIG.  183 


Fig.  182  shows  the  pitch  cylinders  of  a  pair  of  helical  gears.     The 
line  MN  is  the  common 'tangent  to  the  teeth  at  the  point  of  contact  of 


GEARS  AND  GEAR  TEETH  151 

the  pitch  cylinders  (that  is,  the  pitch  point).  B  is  therefore  the  angle 
of  the  helix  of  the  driver  and  BI  the  angle  of  the  helix  of  the  driven 
gear.  Here  the  angle  0  between  the  two  shafts  is  equal  to  180° 

Fig.  183  is  the  development  of  the  surfaces  of  the  two  pitch  cylin- 
ders shown  in  Fig.  182,  the  slanting  lines  being  the  development  of 
imaginary  helices  at  the  centers  of  the  teeth  on  the  pitch  cylinders. 
The  perpendicular  distance  between  these  lines  is  the  normal  pitch  P 
(the  same  in  both  gears).  The  distance  C  and  C\  between  the  points 
of  intersection  of  two  adjacent  teeth  with  the  ends  of  the  cylinder 
(EF  and  EiFi)  are  the  circular  pitches  of  the  respective  gears.  It  will 
be  noticed  that  the  circular  pitches  of  the  two  gears  are  not  alike  but 
depend  upon  the  helix  angles. 

It  should  be  noticed  that  if  the  relation  between  the  angles  B  and  /3  is 
such  that  /Si  becomes  0  the  driven  gear  becomes  like  an  ordinary  spur 
gear.  Hence  a  properly  formed  worm  may  be  made  to  drive  a  spur  gear 
if  their  axes  are  set  at  the  proper  angle  with  each  other. 

165.  Relation  between  the  Circular  Pitches  of  a  Pair  of  Helical 
Gears.  Referring  still  to  Fig.  183 

p  T> 

C  = ~  and  Ci  = 


I-*     c*uv>   v/i  T-> 

cos  B  cos  BI 

mu        £  Ci         COS  B 

Therefore,  ~  = =-  •  (52) 

C        cos  z>i 

166.  Relation  between  Numbers  of  Teeth  in  a  Pair  of  Helical 
Gears.  Let  T  represent  the  number  of  teeth  in  the  driver  (Fig.  183) 
and  Ti  the  number  of  teeth  in  the  driven  gear. 

~T\  _T\ 

Then  T  =  -, 


mr,  i 

Therefore  _  = 

C 
But  from  equation  (52)  ^= 


cos  B 


That  is,  the  numbers  of  teeth  are  inversely  as  the  product  of  the  pitch  diam- 
eters multiplied  by  the  cosines  of  the  helix  angles. 


152  ELEMENTS  OF  MECHANISM 

167.  Speed  Ratio  of  Helical  Gears.  As  in  the  case  of  spur  gears, 
the  angular  speeds  of  a  pair  of  helical  gears  are  inversely  as  the  numbers 
of  teeth.  If  N  represents  the  angular  speed  of  the  driver,  and  Ni  of  the 
driven  gear,  it  follows  from  equation  (53)  that 

Ni=D  GQsB 
N      Dicostfi* 


CHAPTER  VI 
WHEELS   IN   TRAINS 

168.  Train  of  Wheels.     A  train  of  wheels  is  a  series  of  rolling  cylin- 
ders or  cones,  gears,  pulleys  or  other  similar  devices  serving  to  transmit 
power  from  one  shaft  to  another. 

The  examples  of  rolling  cylinders,  gears,  etc.,  which  have  been  dis- 
cussed in  earlier  chapters  are  really  wheel  trains  each  involving  only 
one  pair  of  wheels.  In  Fig.  184  D  is  a  gear  fast  to  shaft  A.  E  is  a  gear 
fast  to  shaft  B  and  meshing  with  D.  F  is  another  gear  also  fast  to  shaft 
B  and  meshing  with  the  gear  G 
which  is  fast  to  shaft  C.  If  now 
the  shaft  A  begins  to  turn,  D  will 
turn  with  it,  and,  therefore,  cause 
E  to  turn.  Since  E  is  fast  to  the 
shaft  B  the  latter  will  turn  with  E. 
Gear  F  will  then  turn  at  the  same 
angular  speed  as  E  and  will  cause  G 
to  turn,  causing  the  shaft  C  to  turn 
with  it.  That  is,  D  drives  E,  and  F, 
turning  with  E,  drives  G. 

The  above  is  an  example  of  a 
simple  train  of  gears  consisting  of 
two  pairs.  Fig.  185  shows  an 
arrangement  of  pulleys  similar  in 
action  to  the  gears  shown  in  Fig. 
184.  H  is  a  pulley  on  the  shaft  R 
belted  to  the  pulley  J  on  shaft  S.  On  the  same  shaft  is  another  pulley 
K  belted  to  the  pulley  L  on  shaft  T. 

Fig.  186  shows  a  train  of  wheels  involving  both  gears  and  pulleys. 
In  this  case  D  is  a  gear  on  shaft  A,  meshing  with  and  driving  the  gear 
E  on  shaft  B.  On  the  same  shaft  is  pulley  K,  belted  to  the  pulley  L  on 
shaft  C. 

169.  Driving  Wheel  and  Driven  Wheel.     Referring  again  to  Fig. 
184,  the  gear  D  by  its  rotation  causes  E  to  turn;  therefore,  D  may  be 
called  the  driver  or  driving  wheel,  and  E  the  driven  wheel.     Similarly  F, 
turning  with  E,  is  the  driver  for  the  wheel  G.     Hence,  in  any  train  such 

153 


FIG.  184 


154 


ELEMENTS  OF  MECHANISM 


as  here  shown,  consisting  of  three  axes  with  two  pairs  of  wheels,  two  of 
the  wheels  are  drivers  and  two  are  driven  wheels. 

170.  Idle  Wheel.     In  Fig.  187,  gear  D  drives  E,  which  in   turn 
drives  F.     E  is,  therefore,  both  a  driven  and  a  driving  wheel.     Such 
a  wheel  is  called  an  idle  wheel. 

171.  Train  Value  (Speed  Ratio).     The  ratio  of  the  angular  speed 
of  the  last  wheel  of  a  train  to  the  angular  speed  of  the  first  wheel  of  the 


FIG.  185 

same  train  is  called  the  value  of  the  train,  or  train  value,  and  will  be  rep- 
resented by  e.  For  example,  if  the  shaft  A  in  Fig.  184  makes  25  r.p.m., 
and  the  sizes  of  the  several  gears  are  such  that  shaft  C  makes  150  r.p.m., 


^ 

L 

1 

1       /                              Ml/ 

1 

| 

FIG.  186 

the  value  of  the  train  would  be  ^ff-  =  6=6.  An  inspection  of  the 
same  figure  will  show  that  if  A  turns  right  handed,  B  will  turn  left 
handed  and  C  will  turn  right  handed.  The  direction,  then,  of  C  is  the 
same  as  that  of  A.  The  value  of  this  train  is  then  said  to  be  positive, 


WHEELS  IN  TRAINS 


155 


and  will  be  indicated  by  putting  a  +  sign  in  front  of  the  number  which 
indicates  its  value.  If  the  number  of  wheels  involved  is  such  that  the 
last  shaft  turns  in  the  opposite  direction  from  the  first  shaft,  the  value 
of  the  train  will  be  said  to  be  negative,  which  fact  will  be  indicated  by 
a  —  sign  in  front  of  the  number  indicating  the  train  value. 

172.  Calculation  of  Speeds.  Let 
it  be  assumed  that  the  gears  in 
Fig.  184  have  teeth  as  follows: 

D,  100  teeth         F,  125  teeth 

E,  50  teeth         G,     25  teeth 

It  will  also  be  assumed  that  shaft 
A  makes  25  r.p.m.  and  it  is  re- 
quired to  find  the  speed  of  C. 
Since  the  speed  of  B  is  to  the 
speed  of  A  as  the  teeth  in  D  are  to  the  teeth  in  E,  the  revolutions  of  B 
will  be  equal  to  25  X  *££• ;  also,  since  the  speed  of  C  is  to  the  speed  of  E 
as  the  teeth  in  F  are  to  the  teeth  in  G,  the  speed  of  C  =  25  X 
X  *££•  =  250.  Expressing  this  as  a  formula, 

The  speed  of  the  last  shaft  is  equal  to  the  speed  of  the  first  shaft 
the  product  of  the  teeth  of  all  the  drivers 


FIG.  187 


X 


the  product  of  the  teeth  of  all  the  driven  wheels 


(55) 


In  the  case  of  pulleys  in  Fig.  185  the  principle  is  the  same,  except 
that  diameters  are  used  instead  of  numbers  of  teeth.  Suppose  that 
pulley  H  is  24  ins.  in  diameter,  J  8  ins.  in  diameter,  K  36  ins.  in  diameter, 
and  L  12  ins.  in  diameter,  then  the  speed  of  T  will  be  equal  to  the  speed 

24  X  36    ,, 

;  that  is,  in  the  case  of  a  train  of  pulleys: 


of 


X 


8  X  12 


The  speed  of  the  last  shaft  is  equal  to  the  speed  of  the  first  shaft 
the  product  of  the  diameter  of  all  the  driving  pulleys 
the  product  of  the  diameter  of  all  the  driven  pulleys 


X 


(56) 


In  a  train  consisting  of  a  combination  of  gears  and  pulleys,  as  in  Fig. 
186, 

The  speed  of  the  last  shaft  is  equal  to  the  speed  of  the  first  shaft 
the  product  of  diameters  and  numbers  of  teeth  of  all  the  driving  wheels 


X 


the  product  of  diameters  and  numbers  of  teeth  of  the  driven  wheels 


(57) 


An  idle  wheel  such  as  gear  E  in  Fig.  187  has  no  effect  on  the  speed 
ratio,  but  does  cause  a  change  in  the  direction.  This  can  be  seen  from 
the  following  calculation:  Let  the  wheel  D  have  100  teeth;  E  75  teeth, 


156 


ELEMENTS  OF  MECHANISM 


FIG.  188 


and  F  25  teeth,  then  the  speed  of  shaft  C  is  equal  to  the  speed 
of  A  X  J^  X  -Jf  .  75  then,  which  is  the  number  of  teeth  in  the  idle 
wheel,  appears  in  both  numerator  and  denominator  and  cancels  out, 

and  therefore  the  speed  of  C  be- 
comes the  speed  of  A  X  *££-. 

173.  Driving  and  Driven  Gears 
having  Coincident  Axes.  Fig.  188 
is  a  diagram  of  the  back  gear 
arrangement  for  a  simple  cone 
pulley  head-stock  on  an  engine 
lathe.  It  illustrates  the  principles 
involved  when  two  wheels,  whose 
axes  coincide,  are  connected  by  a 
train  of  wheels  through  an  inter- 
mediate shaft,  the  axis  of  the 
intermediate  shaft  being  parallel 
to  the  axis  of  the  connected  wheels.  P  is  the  cone  pulley.  A  is  a 
gear  integral  with  P  and  meshing  with  gear  B.  C  is  another  gear  on 
the  same  shaft  with  B,  both  B  and  C  being  fast  to  the  shaft.  C  meshes 
with  gear  D  on  the  spindle  S.  From  Eq.  55. 
Speed  of  spindle  =  speed  of  cone  pulley 

Teeth  in  A  X  Teeth  in  C 
Teeth  in  B  X  Teeth  in  D 

Since,  however,  the  shaft  R  is  parallel  to  S  the  gears  must  be  so  pro- 
portioned that  the  pitch  radius  of  A  +  pitch  radius  of  B  equals  pitch 
radius  of  C  +  pitch  radius  of  D.  Consequently,  if  the  pitches  of  the 
two  pairs  are  to  be  in  some  definite  ratio  there  must  be  a  corresponding 
relation  between  the  sum  of  the  teeth  in  A  and  B  and  the  sum  of  the 
teeth  in  C  and  D. 

174.  Frequency  of  Contact  between  Teeth:  Hunting  Cog.  Given 
two  gears  Gi  and  Gz  with  teeth  Ti  and  T%  having  the  greatest  common 
divisor  a. 

Then  let  Ti  =  ati  and   T*  =  ah. 

rp,       ,  Turns  ft       Tz      ok      fe 

Theref°re>  Tu^ft  =  7\=^=F; 

Contact  between  the  same  pair  of  teeth  will  take  place  after  the  passage  of 
a  number  of  teeth  equal  to  the  least  common  multiple  of  TI  and  T2=atitz. 
Therefore,  before  the  same  pair  of  teeth  come  in  contact  again  after 
having  been  in  contact, 

the  turns  of  ft  = 


and  turns  of  ft  =  —r~  = 


WHEELS  IN  TRAINS  157 

Therefore  the  smaller  the  numbers  ti  and  fe,  which  express  the  velocity 
ratio  of  the  two  axes,  the  more  frequently  will  the  same  pair  of  teeth  be 
in  contact. 

Assume  the  velocity  of  ratio  of  two  axes  to  be  nearly  as  5  to  2.  Now, 
if  Ti  =  80  and  T2  =  32, 


and  the  same  pair  of  teeth  will  be  in  contact  after  five  turns  of  Tz  or 
two  turns  of  TI. 

If  Ti  is  changed  to  81,  then  ^  =  -i  =  ->  very  nearly,  the  angular 

L  2         O&          £ 

velocity  ratio  being  scarcely  distinguishable  from  what  it  was  originally. 
But  now  the  same  teeth  will  be  in  contact  only  after  81  turns  of  TV  or 
32  turns  of  7\. 

The  insertion  of  a  tooth  in  this  manner  prevents  contact  between  the 
same  pair  of  teeth  too  often,  and  insures  greater  regularity  in  the  wear 
of  the  wheels.  The  tooth  inserted  is  called  a  hunting  cog,  because  a 
pair  of  teeth,  after  being  once  in  contact,  gradually  separate  and  then 
approach  by  one  tooth  in  each  turn,  and  thus  appear  to  hunt  each  other 
as  they  go  round.  In  cast  gears,  which  will  be  more  or  less  imperfect, 
it  would  be  much  better  if  an  imperfection  on  any  tooth  could  distribute 
its  effect  over  many  teeth  rather  than  that  all  the  wear  due  to  such 
imperfection  should  come  always  upon  the  same  tooth.  This  result  is 
most  completely  obtained  when  the  numbers  of  teeth  on  the  two  gears 
are  prime  to  each  other,  as  above,  when  T\  and  T2  were  81  and  32 
respectively. 

175.  Example  of  Wheels  in  Trains.     The  following  paragraphs  will 
give  a  few  examples  of  wheels  in  trains.     These  are  selected  because 
they  serve  to  illustrate  the  principles  involved,  and  not  because  a  knowl- 
edge of  these  particular  trains  is  of  special  importance. 

176.  Clockwork.     A  familiar  example  of  the  employment  of  wheels 
in  trains  is  seen  in  clockwork.     Fig.  189  represents  the  trains  of  a  com- 
mon clock;  the  numbers  near  the  different  wheels  denote  the  number 
of  teeth  on  the  wheels  near  which  they  are  placed. 

The  verge  or  anchor  0  vibrates  with  the  pendulum  P,  and  if  the 
pendulum  vibrates  once  per  second,  it  will  let  one  tooth  of  the  escape- 
wheel  pass  for  every  double  vibration,  or  every  two  seconds.  Thus  the 
shaft  A  will  revolve  once  per  minute,  and  is  suited  to  carry  the  second 
hand  S. 

The  value  of  the  train  between  the  axes  A   and  C  is   -  -  T 

8X8         1  S 

=  6Q  x  64  =  QQ  or  tne  shaft  C  revolves  once  for  sixty  revolutions  of 


158 


ELEMENTS  OF  MECHANISM 


A;  it  is  therefore  suited  to  carry  the  minute  hand  M.  The  hour  hand 
H  is  also  placed  on  this  shaft  C,  but  is  attached  to  the  loose  wheel  F  by 
means  of  a  hollow  hub.  This  wheel  is  connected  to  the  shaft  C  by 
means  of  a  train  and  intermediate  shaft  E.  The  value  of  this  train  is 
turns  H  _  28  X  8  1 
turns  M  ~  42  X  64  ~~  12* 

The  drum  D,  on  which  the  weight-cord  is  wound,  makes  one  revolu- 
tion for  every  twelve  of  the  minute  hand  M,  and  thus  revolves  twice 

each  day.  Then,  if  the  clock  is  to  run 
eight  days,  the  drum  must  be  large  enough 
for  sixteen  coils  of  the  cord.  The  drum 
is  connected  to  the  wheel  G  by  means  of 
a  ratchet  and  click,  so  that  the  cord  can 
be  wound  upon  the  drum  without  turning 
the  wheel. 

Clock  trains  are  usually  arranged  as 
shown  in  the  figure,  the  wheels  being  placed 
on  shafts,  often  called  "  arbors,"  whose 
bearings  are  arranged  in  two  parallel  plates 
which  are  kept  the  proper  distance  apart 
by  shouldered  pillars  (not  shown)  placed 
at  the  corners  of  the  plates.  When  the 
arbor  E  is  placed  outside,  as  shown,  a 
separate  bearing  is  provided  for  its  outer 
end. 

177.  Planer  Drive.  Fig.  190  shows  a 
portion  of  the  gear  train  for  driving  the 
platen  on  a  planer.  The  shaft  S  is  driven 
by  a  motor  or  by  a  counter-shaft  above 
the  machine.  S  drives  Si  by  the  pair  of 
gears  A  and  B.  On  Si  are  two  gears  E 
and  D  which  are  fast  to  each  other  and 
slide  on  a  long  key  on  the  shaft  so  that 
they  are  obliged  to  turn  with  the  shaft,  but 
may  be  moved  along  it  by  means  of  a 
shifting  device  fitting  into  the  groove  V. 
A  similarly  arranged  pair  of  sliding  gears 
H  and  K  are  on  the  shaft  S2.  Gears  B 
and  C  are  permanently  attached  to  Si  and 

F  and  G  to  Sz.  The  shaft  S2  has  a  pulley  on  the  end  (not  shown  in 
the  diagram)  which  drives,  by  a  belt,  to  the  mechanism  operating  the 
table.  The  object  of  the  system  of  gears  shown  in  the  diagram  is  to 


0 

: 

8t 

—  —  i          j 

3ft 

j 

60  t 

i 

9f 

£4* 

28  t 

54  f 

£ 

8t 

" 

- 

- 

42  f 

L 

£ 

G 

W 

y 

96  1 

FIG.  189 

M 


H 


WHEELS  IN  TRAINS 


159 


enable  the  operator  to  obtain  four  different  speeds  for  &,  and  therefore 
for  the  platen,  for  one  speed  of  S. 

With  H  and  K  in  the  position  shown  and  with  D  and  E  to  the  left  so 
that  E  meshes  with  F  as  shown,  E  is  driving  F  and  the  train  value 

between  S  and  S2  is 

Teeth  A  X  Teeth  E 

Teeth  B  X  Teeth  F 

All  the  other  gears  are  turning  idle.     If  D  and  E  are  moved  to  the  right 
a  distance  just  a  little  more  than  the  width  of  the  face  of  E  this  gear 


-O 


/>/V/V?0 


FIG.  190 

will  be  out  of  mesh  with  F  and  the  whole  system  will  be  in  neutral  posi- 
tion.    If  D  and  E  are  moved  still  further  to  the  right  D  will  mesh  with 

A      D 

G  and  the  train  from  S  to  Sz  is  -5  X  ~  • 

JD       Or 

Now,  putting  D  and  #  back  into  neutral,  slide  H  and  K  to  the  right 


until  K  meshes  with  B  and  the  train  between  S  and  $2  is  •_  X      -     If 
#  and 


2       _ 
are  slid  to  the  left  H  will  mesh  with  C  and  the  train  will  be 


B      H 

The  levers  which  operate  the  two  sliding  units  are  interlocked  in 
such  a  way  that  one  of  the  units  must  be  in  neutral  position  before  the 
other  can  be  moved  to  the  right  or  left  of  neutral. 


160 


ELEMENTS  OF  MECHANISM 


178.  Automobile  Transmission.  Fig.  191  is  a  diagram  of  the  ar- 
rangement of  the  gears  in  a  common  type  of  automobile  transmission 
which  allows  three  speeds  forward  and  one  reverse. 

The  gear  A  is  on  the  end  of  a  sleeve  driven  directly  from  the  motor. 
A  turns  freely  on  the  end  of  the  propeller  shaft  P.  On  the  counter- 
shaft S  are  four  gears,  B,  D,  F,  and  G  which  are  attached  to  each  other 
and  turn  as  a  unit.  The  purpose  of  the  system  of  gears  is  to  make  it 
possible  for  A,  turning  at  a  definite  speed,  to  drive  shaft  P  at  three 


FIG.  191 

different  speeds  in  the  same  direction  that  A  itself  is  turning,  and  at 
one  speed  in  the  direction  opposite  that  of  A. 

B  is  in  mesh  with  A  and  the  counter-shaft  unit  is  turning  in  a  direc- 
tion opposite  that  of  A  at  a  speed  which  is  to  the  speed  of  A  equal  to 

™ — 7T— -•  —5 '     With  gears  C  and  E  in  the  positions  shown,  the  counter- 
Teeth  in  B 

shaft  unit  is  turning  idle  and  the  whole  system  is  in  neutral.     On  the 
left  end  of  A  is  the  hub  M  which  has  teeth  on  its  circumference.     In 


WHEELS  IN  TRAINS 


161 


the  right  end  of  gear  C  is  a  cavity  with  spaces  into  which  the  teeth  of 
M  can  lock  if  C  is  moved  to  the  right.  C  slides  on  a  key,  or  keys,  in  shaft 
P  so  that  when  C  turns  P  turns  with  it.  If  the  gear-shifting  lever  is 
moved  to  the  proper  position  C  will  be  pushed  over  M  thus  locking  C 
to  A  and  the  shaft  P  will  be  driven  at  the  same  speed  as  A.  This  puts 
the  transmission  in  "high."  If  C  is  pushed  to  the  left  of  its  present 
position  it  will  mesh  with  D  and  then 

Speed  P  =  Teeth  in  A  X  Teeth  in  D 
Speed  A  ~  Teeth  in  B  X  Teeth  in  C ' 

This  gives  the  "second"  or  "intermediate"  speed.  If  C  is  put  in  the 
position  shown  and  the  shifting  lever  moved  so  as  to  take  hold  of  gear 
E  and  move  it  to  the  right  to  mesh  with  F 
the  speed  ratio  will  be  

Speed  P  _  Teeth  in  A  X  Teeth  in  F  ~*  A    ^" ' 

Speed  A  ~  Teeth  in  B  X  Teeth  in  E ' 


This  gives  the  "third"  or  "low"    speed. 
Pushing  E  to  the  left  of  its  neutral  posi- 
tion brings  it  into  mesh  with  the  idler  K  - 
which  is  driven  from  G  giving  a  speed 
ratio 

Speed  P  =  TeethinA  X  Teething 
Speed  A      Teeth  in  B  X  Teeth  in  E ' 


130 1 

37 1 
35 1 
31 1 
26 1 


33 1 


IJt 


FIG.  192 


and  since  the  connection  between  G  and 
E  is  through  the  idler,  the   direction  of  - 
E  and   therefore   of  P  is  opposite  that 
of  A,  giving  the  "  reverse  "  drive. 

179.   Cotton  Card  Train.     Fig.  192  shows  the  train  in  a  cotton  card- 
ing-machine.     For  the  train  we  have  the  value 
turns  B  _  135      37       130      17 
turns  A  ~  17       20      W  X  33  "      •<*7-o4- 

In  the  machine  the  lap  of  cotton  passing  under  the  roll  A  is  much  drawn 
out  in  its  passage  through  the  machine,  and  it  becomes  necessary  to 
solve  for  the  ratio  of  the  surface  speeds  of  the  rolls  B  and  A.  For  this, 
since  the  surface  speed  equals  2  ir  X  turns  X  radius  or  turns  X  TT  X 
diameter, 

surface  speed  B  _  turns  B  X  diam.  B 
surface  speed  A  ~  turns  A  X  diam.  A ' 

4 


surface  speed  B  = 

surface  speed  A      2.25 


=  67.27. 


162 


ELEMENTS  OF  MECHANISM 


180.  Hoisting  Machine  Train.  A  train  of  spur  gears  is  often  used 
in  machines  for  hoisting  where  the  problem  would  be  to  find  the  ratio 
of  the  weight  lifted  to  the  force  applied.  In  Fig.  193  the  value  of  the 

train  is 

turns  B  _ 
turns  A 
D  =  15" 


100 1 


2jsi    then,  if 


84 1 


"84"  16' 
R  =  li  ft., 

X15_.i_. 
A  30      32' 


21 

:  100 

and 

speed  W      JL_ 
speed  F  ~  16 

F_  _  speed  W      ^_ 
"    W  ~  speed  F  ~  32 ' 

or  if  F  were  50  Ibs.,  W  would  be  1600  Ibs. 

if  loss  due  to  friction  were  neglected. 

181.  Reduction  Gear.  Fig.  194,  furnished  by  the  Fellows  Gear 
Shaper  Co.,  shows  a  train  of  gears  consisting  of  a  pinion  concentric  with 
an  annular,  with  three  idle  gears.  The  gears  in  this  case  are  herring- 
bone gears. 


W 
FIG.  193 


FIG.  194 


182.  Designing  Gear  Trains.  No  definite  rules  or  formulas  can  be 
followed  in  designing  a  train  of  gears  to  have  a  certain  train  value.  The 
process  is  mainly  one  of  "  cut  and  try  "  until  the  desired  result  is  obtained. 
There  are,  however,  some  general  lines  of  attack  which  may  be  followed, 


WHEELS  IN  TRAINS 


163 


and  which  may  best  be  understood  by  studying  certain  typical  prob- 
lems. If  the  value  of  the  train  is  chosen  arbitrarily,  it  may  be  found 
impossible  to  select  gears  which  will  give  exactly  the  value  called  for. 

Example  24.  Let  it  be  required  to  select  the  gears  for  a  train  in  which  the  last 
gear  shall  turn  19  times  while  the  first  gear  turns  once,  the  direction  of  rotation  being 
the  same.  No  gear  is  to  have  less  than  12  teeth  nor  more  than  60  teeth. 

Solution.     The  first  step  in  I  , 

the  solution  of  this  problem  is 
the  determination  of  the  num- 
ber of  pairs  of  gears  necessary 
to  give  a  train  value  of  19  and 
keep  the  gears  within  the  limits 
of  size  specified.  If  only  one 
pair  were  used,  making  the 
driver  as  large  as  allowed,  that 
is,  60  teeth,  and  the  driven 
gear  as  small  as  allowed,  that 
is,  12  teeth,  the  train  value 
would  be  ff  or  5.  If  a  second 
driver  of  60  teeth  is  made  fast 

to  the  12  teeth  gear,  and  this  drives  a  second  gear  of  12  teeth,  as  shown  in  Fig.  195, 
the  train  value  becomes  f  I  X  f  §  =  25.  This  is  greater  than  the  assigned  value  of 
19,  therefore,  two  pairs  of  gears  will  be  sufficient  to  obtain  a  value  of  19  without 
exceeding  the  specified  limits  of  size.  Having  thus  determined  the  number  of  gears 
necessary,  the  next  step  is  the  selection  of  the  gears  themselves  to  give  the  exact 
value  of  19.  This  may  be  tried  in  several  ways.  First,  since  two  pairs  of  gears  are 

to  be  used,  the  square  root  of  19 
may  be  taken.  This  is  approxi- 
mately 4.36.  This  does  not  differ 
greatly  from  4f .  Now 

a  X!i9  _  19 
X 


FIG.  195 


Multiply  both  numerator  and 
denominator  of  the  first  fraction 
by  12,  and  of  the  second  fraction 
by  3  and  the  equation  becomes 

FIG.  196  H  X  «  -  19. 

Therefore,   a    train    of    gears    as 

shown  in  Fig.  196  will  give  the  required  train  value  of  19.     If  the  directional  relation 
were  not  correct  an  idler  might  be  used. 

Example  25.  Let  it  be  required  to  find  suitable  gears  to  drive  a  shaft  at  4  r.p.m. 
from  a  shaft  making  500  r.p.m.,  using  no  gear  of  less  than  24  teeth  nor  more  than  96 
teeth. 

Solution.    The  train  value  in  this  case  is  -%fa  =  Tib- 

Since  the  smallest  gear  that  can  be  used  is  24  teeth  and  the  largest  96  teeth,  the 


164 


ELEMENTS  OF  MECHANISM 


Therefore  i  must  be  multiplied  by  itself  a  sufficient  number  of  times  to  obtain 
a  final  result  less  than  T£ 5.  Now,  (|)3  =  g'j;  then  3  pairs  of  gears  will  not  be  enough. 
But  (|)4  =-5%%  ,  which  is  beyond  the  specified  T£S,  hence  4  pairs  will  be  proper  to  use. 

1  3 

The  fourth  root  of  ^-g  is  approximately  ^r  or  ^r- 

OgT  1U 

A  train  of  gears  for  a  trial  might  be  in  the  ratio  of 


3_ 
10 


0  q 

—  V  — 

10  X  10 


1XJ.X1X! 

10      10      10      27' 


Multiplying  numerator  and  denominator  of  each  of  the  first  three  terms  by  9  and  of 
the  last  term  by  3  gives. 

IS  x  13  x  f $  x  If. 

Then  a  train,  such  as  shown  in  Fig.  150,  fulfils  the  required  conditions. 


FIG.  197 

Example  26.    Where  an  error  of  a  certain  amount  is  allowable,  as  would  very 
often  be  the  case,  the  following  method  may  be  used  to  advantage. 


Let  the  value  of  the  train  be  60  and  — 


100 
20 


where  T  represents  the  maximum  number  of  teeth  and  t  the  minimum.  It  will  be 
found  that  three  pairs  of  gears  are  needed.  Therefore  take  the  cube  root  of  60, 
which  is  3.91 +,  and  write 

3.91 


1 


x 


x 


1 


Since  the  small  gears  are  not  to  have  less  than  20  teeth,  and  since  20  X  3.91  =  78+, 
a  first  approximation  may  be  written 

f  §  X  £8  X  £8, 

which  will  be  found  to  equal  61.63;  if  this  result  is  too  greatly  in  error,  a  reduction 
of  one  or  two  teeth  in  the  numerator  or  an  increase  in  the  denominator  may  give  a 
closer  result,  as 

IS  X  18  X  18  =  60.07. 


WHEELS  IN  TRAINS  165 

Example  27.  To  design  a  train  of  four  gears,  with  the  axis  of  the  last  wheel  coin- 
cident with  the  axis  of  the  first  wheel  as  in  Fig.  188.  The  train  value  to  be  TV  No 
gear  to  have  less  than  12  teeth.  All  gears  to  be  of  the  same  pitch. 

Solution.  Since  there  are  two  pairs  in  the  train,  the  value  ^  must  be  separated 
into  two  factors  and  it  is  desirable  to  have  these  factors  as  nearly  as  may  be  of  the 
same  value.  The  square  root  of  ^  is  between  |  and  |  so  a  trial  pair  of  factors  may 
be  taken  J  X  i-  Then  letting  the  letters  Ta,  T&,  Tc,  Ta  represent  the  numbers  of 
teeth  in  the  gears  A,  B,  C,  D  respectively  (Fig.  188). 


Now,  since  the  pitches  are  all  alike, 

Ta  +  Tb  =  Tc  +  Ta  (see  §  173). 

Let  Z  represent  this  sum.  Then  a  value  must  be  chosen  for  Z  such  that  it  may  be 
broken  up  into  two  parts  whose  ratio  is  |  and  also  two  parts  whose  ratio  is  |. 

If  Z  is  made  equal  to  the  least  common  multiple  of  1  +  3  and  1+4,  the  condi- 

tion will  be  satisfied.  This  L.  C.  M  .  is  20.  Then  ^  would  be  ^  and  ^  would  be  4'  ' 

1  6  15  Id  16 

But  these  values  are  too  small  for  the  numbers  of  teeth  in  the  gears.  Then 
numerator  and  denominator  of  both  fractions  must  be  multiplied  by  some  number 
such  that  no  number  will  be  less  than  the  number  of  teeth  allowed  in  the  smallest 
gear.  In  this  case  multiplying  T\  and  T4ff  each  by  f  gives  if  and  if. 

Therefore        Ta  may  be  15,       Ti,  =  45,       Tc  =  12,       Td  =  48. 

The  above  method  (Example  27)  may  be  expressed  as  follows: 
When  both  pairs  have  the  same  pitch. 

If  7  X  7  are  the  factors  of  e  (i.e.,  the  train  value)  expressed  in  lowest 

t2  &4 

terms,  then  Ta  +  !F6  and  Tc  +  T&  must  be  made  equal  to  the  L.  C.  M.  of 
ti  +  tz  and  Z3  +  U  or  to  some  multiple  of  the  L.  C.  M. 

When  the  pitches  of  the  two  pairs  are  different. 

The  case  illustrated  in  Example  27  is  not  a  practical  one,  because  the 
stresses  on  the  second  pair  of  gears  are  always  greater,  requiring  a 
greater  circular  pitch. 

If  the  pitch  number  of  A  and  B  =  PI  and  the  pitch  number  of  C  and 

D  =  P2,  and  if  —  •  =  -^  reduced  to  its  lowest  terms,  then  Ta  +  Td  is 

PZ          Z2 

made  equal  to  the  L.  C.  M.  of  ti  +  12  and  U  +  U  (or  to  some  multiple 
of  the  L.  C.  M)  multiplied  by  pi,  and  Tc+Td  is  made  equal  to  the 
L.  C.  M.  (or  the  same  multiple  of  the  L.  C.  M.)  multiplied  by  pz. 


CHAPTER  VII 


EPICYCLIC   GEAR  TRAINS 

183.  An  epicyclic  train  of  gears  is  a  train  in  which  some  of  the 
gears  turn  on  fixed  axes,  while  others  turn  on  axes  which  are  them- 
selves in  motion.  The  wheels  are  usually  connected  by  a  rigid  link 
known  as  the  train  arm  which  rotates  on  the  axis  of  one  of  the 
wheels  of  the  train. 

Assume  that  C,  Fig.  198,  is  a  gear  carried  by  the  arm  A  and  pinned 
to  A  so  that  it  cannot  turn  on  its  own  axis.  If  A  is  caused  to  turn 
about  S  once,  a  reference  mark  V  on  C,  which  in  the  position  shown  is 
pointing  downward,  would  point  in  every  direction  successively  as  A 

revolved  and  come  finally  to  its 
present  position  when  the  arm  had 
made  a  complete  turn. 

If  C  is  meshing  with  another  gear 
on  the  axis  S,  and  therefore  turns 
on  its  axis  at  the  same  time  that  the 
axis  revolves,  the  reference  mark 
will  swing  around  and  point  to  its 
original  direction  a  number  of  times 
equal  to  the  algebraic  sum  of  tthe 
speed  of  the  arm  and  the  speed  of 
C  on  its  axis.  This  resultant  number  of  times  that  the  reference  mark 
returns  to  its  original  direction,  having  always  turned  completely 
over,  is  called  the  number  of  absolute  turns  that  C  makes,  or  its  absolute 
speed.  The  speed  of  C  on  its  own  axis  is  called  its  relative  speed,  or  speed 
relative  to  the  arm.  Either  direction  of  rotation  may  be  assumed  as 
positive  (+) ;  then  rotation  in  the  opposite  direction  must  be  considered 
as  negative  (  — ). 

Fig.  199  illustrates  an  epicyclic  train,  and  the  following  description 
of  its  operation  should  be  studied  carefully  in  order  to  understand  the 
principle  of  action.  When  this  principle  is  clear  the  analysis  of  any 
epicyclic  train  is  comparatively  simple. 

B  is  a  gear  turning  with  the  shaft  S  which  is  in  stationary  bearings 
and  is  driven  by  the  gears  R  and  K.  C  is  a  gear  meshing  with  B.  The 
stud  T  on  which  C  turns  is  carried  by  the  arm  A.  Fast  to  the  hub  of 
A  is  the  gear  E,  driven  by  the  gear  D.  Attached  to  C  is  the  gear  F 
which  drives  G.  G  turns  on  the  same  axis  as  B,  but  must,  of  course, 
be  free  to  turn  at  a  different  speed  from  B. 

166 


FIG.  198 


EPICYCLIC  GEAR  TRAINS 


167 


D  and  R  receive  their  motion  from  outside  sources.  The  resultant 
speed  of  G,  due  to  the  combined  speeds  of  B  and  A,  is  the  algebraic 
sum  of  the  speeds  which  it  would  have  when  each  moved  with  the  other 
standing  still. 

If  the  gear  D  is  first  assumed  not  to  turn,  the  arm  A  will  be  station- 
ary, and  the  following  equation  will  hold  true 

Speed  of  G  =  Teeth  in  B  X  Teeth  in  F, 
Speed  of  B  ~ 

(I) 


or 


Teeth  in  C  X  Teeth  in  G 
Speed  of  G  =  speed  of  B  X  train  value. 


\ 

/>  

^^"fE 

8\ 

•H-                        i  T 

0                                              '1                        n 
\                                          II                    -G 

i  * 

\      !   !     *- 

_H  i.l.i 

+1  ,    \  ^ 

1  > 

i 

B       l      vj    '\ 

I 

X      A         V. 

i  i  , 

kl  v          T 

v  71 

\ 

R 

FIG.  199 

If,  on  the  other  hand,  the  gear  B  is  assumed  not  to  turn  and  D  turns 
at  a  definite  speed,  the  arm  A  will  revolve,  the  stud  T  will  travel  around 
S  as  an  axis,  C  rolling  around  on  B,  and  F  rolling  around  on  G.  This 
will  impart  motion  to  G  which  may  be  found  as  follows :  Suppose  A  to 
have  a  speed  of  a  r.p.m.  right-handed;  then  the  speed  of  C  on  its  own 
axis  is  the  same  as  if  A  were  held  still  and  B  turned  with  a  r.p.m.  left- 
handed.  That  is,  C  would  have  a  speed  on  its  axis  of  a  X  ™ — -r—. — ~ 

leetn  in  (_/ 

r.p.m.  right-handed,  relative  to  its  own  axis,  or  relative  to  the  arm. 
This  speed  of  C  would  impart  to  G,  relative  to  the  arm,  a  speed  of 

Teeth  in  B  X  Teeth  in  F  ^    .  ,  ,,  ,       ,    , 

a  X  ^    xu  .    ^  „  ^    .,   .    „  r.p.m.  =  a  X  Tram  value,  left-handed. 
Teeth  in  C  X  Teeth  in  G      • 


168  ELEMENTS   OF  MECHANISM 

But  the  arm  is  itself  turning  right-handed  at  a  speed  a  r.p.m.   Therefore, 
the  actual  speed  of  G  due  to  the  speed  of  A  is 

a  —  a  X  train  value.  (II) 

Combining  (I)  and  (II) 

Speed  ofG  =  Speed  of  B  X  Train  Value  +  Speed  of  Arm 

—  Speed  of  Arm  X  Train  Value.  (58) 

If  n  represents  the  absolute  turns  or  speed  of  the  last  wheel  of  an 
epicyclic  train  (in  this  case  G),  m  the  absolute  turns  or  speed  of  the 
first  wheel  (B),  a  the  turns  or  speed  of  the  arm  (A),  and  e  the  train 
value,  Equation  (58)  may  be  expressed  thus 

n  =  me  +  a  —  ae.  (59) 

Equation  (59)  is  commonly  written 

n  —  a 


m  —  a 

and,  in  this  form,  may  be  expressed  in  words  thus: 
Turns  last  wheel  relative  to  arm 


(60) 


Train  value  = 


Turns  first  wheel  relative  to  arm 

Absolute  turns  of  last  wheel  —  turns  of  arm 

Absolute  turns  of  first  wheel  —  turns  of  arm 


Problems  relating  to  epicyclic  trains  may  be  solved  by  the  formula 
given  in  Equation  (59)  or  (60)  or  by  another  method  known  as  the 
tabulation  method.  This  consists  in  assuming  that  the  motions  of 
B  and  the  arm  (Fig.  199)  take  place  successively  instead  of  simul- 
taneously. The  gears  are  first  assumed  to  be  made  fast  to  the  arm  so 
that  there  can  be  no  relative  motion.  The  arm  is  made  to  turn  at  the 
proper  speed  for  a  unit  of  time  in  the  proper  direction;  all  the  gears 
will  turn  with  it.  Then  the  arm  is  held  still  and  one  of  the  gears  (in 
this  case  B)  is  turned  backward  or  forward  enough  to  make  its  net 
number  of  turns  equal  to  its  known  speed.  The  sum  of  the  results 
produced  by  these  two  processes  gives  the  net  motion  or  speed  of  the 
other  gear  or  gears. 

If  m,  n,  a,  and  e  have  the  same  meanings  as  in  Equations  (59)  and 
(60)  the  above  process  may  be  tabulated  thus: 

Turns  of  Arm    Turns  of  B  Turns  of  G 

1°  Train  locked a  a  a 

2°  Arm  fixed _JO m  -  a  (m  -  a)  e 

3°  Resultant  motions..  a  m  (m  -  a)  e  +  a 


EPICYCLIC  GEAR  TRAINS  169 

If  the  resultant  number  of  turns  of  G  thus  obtained  be  equated  to  n 
the  resulting  equation  is 

n  =  (m  —  a)  e  +  a 

n  —  a 

or  e  = > 

m  —  a 

which  is  the  same  as  Equation  (60). 

It  is  absolutely  essential  that  the  -\-or-  sign  precedes  each  of  the 
quantities  according  as  its  value  is  positive  or  negative. 

In  applying  either  the  formula  or  the  tabulation  method  care  must 
be  taken  to  include  only  those  gears  which  are  a  part  of  the  epicyclic 
train.  For  instance,  in  Fig.  199  the  gears  D,  E,  R  and  K  serve  merely 
as  drivers  of  members  of  the  epicyclic  train  and  do  not  enter  into  the 
formula  or  the  tabulation. 

184.  Solution  of  Problems  on  Epicyclic  Trains.  The  following  ex- 
amples will  illustrate  the  application  of  the  two  methods  of  solving 
epicyclic  trains.  In  some  cases  the  formula  is  used  and  in  other  cases 
the  tabulation  method,  while  in  a  few  examples  both  methods  are  used. 
Either  method  will  apply  to  any  problem  and  it  is  often  desirable  to 
solve  by  both  for  the  purpose  of  checking. 

Example  28.  In  Fig.  199  let  B  have  80  teeth,  C  40  teeth,  F  90  teeth  and  G  30 
teeth.  If  B  has  a  speed  of  100  r.p.m.  right-handed  and  A  60  r.p.m.  left-handed, 
let  it  be  required  to  find  the  speed  of  G. 

Solution  No.  1.     Using  Eq.  (59)  let  right-handed  rotation  be  assumed  plus.    Then 
m  =  +100,  a  =  -60,  e  =  f£  X  f£  =  6,  and  since  with  the  arm  at  rest  G  would 
turn  in  the  same  direction  as  B,  the  value  of  e  is  plus.     That  is,  e  =  +6. 
Substituting  in  Eq.  (59), 

n  =  100  X  6  +  (-60)  -  (-60  X  6)  =  600  -  60  +  360  =  900. 
Solution  No.  2.     Using  the  tabulation  method. 

Arm  B  G 

Gears  locked  to  arm —  60       —    60  —    60 

Arm  held  still 0       +  160      160  X      6 

-  60       +  100      960  -    60 
=  '900 

Example  29.  In  Fig.  200  let  gear  B  have  24  teeth  and  C  18  teeth.  If  B  is  held 
from  turning  and  the  arm  makes  1  turn  right-handed,  let  it  be  required  to  find  how 
many  absolute  turns  C  makes. 

Solution  No.  1.  Using  Eq.  59,  m  =  0,  e  =  —  f I  =  —  ti  a  =  +1  (assuming 
right-hand  rotation  is  plus).  Then,  substituting  in  Eq.  (59),  n  =  0  +  1  —  (— f), 
or,  n  =  f . 

Solution  No.  2.     Using  the  tabular  method: 

Arm  B  C 

Gears  locked  to  arm +1  +1  +1 

Arm  held  still 0  -  1       -  1  (-  f) 

+  1  0  +  £ 


170 


ELEMENTS  OF   MECHANISM 


Arm 


Example  30.  In  Fig.  201  E  is  an  annular  gear  which  cannot  turn,  being  fast  to 
the  frame  of  the  machine.  The  arm  A  turns  about  the  shaft  S  which  is  also  the 
axis  of  the  gears  B  and  E.  B  has  24  teeth,  C  20  teeth,  D  16  teeth,  and  E  96  teeth. 
Let  it  be  required  to  find  the  speed  of  the  arm  A  to  cause  the  gear  B  to  have  a  speed 
of  75  r.p.m.  left-handed. 

Solution.  Assume  B  to  be  the  first  wheel  of  the  train  and  assume  right-handed 

rotation  as  + : 

Then  referring  to  Eq.  (59)  n  =  0,  m  = 
-75,  e  =  +U  =  +i- 

Substituting  these  values  in  the 
equation, 

0-75X|+a-aXi, 
whence 

a  =  +25. 

Therefore,  A  will  have  to  have  a  speed  of 
25  r.p.m.  right-handed  to  give  the  required 
speed  to  B. 


FIG.  200 


Example  31.  Sun  and  Planet  Wheel.  Fig.  202  shows  an  applica- 
tion of  the  two-wheel  epicyclic  train  known  as  the  Sun  and  Planet 
Wheels,  first  devised  by  James  Watt  to  avoid  the  use  of  a  crank, 
which  was  patented.  In  his  device  the  epicyclic  train  arm  was  replaced 
by  the  stationary  groove  (7,  which  kept  the  two  wheels  in  gear,  a 
represents  the  engine  shaft,  to 
which  the  gear  -D  was  made  fast, 
B  the  connecting  rod,  attached 
to  the  walking  beam.  The  gear 
C  was  rigidly  attached  to  the  end 
of  the  connecting  rod.  While 
with  such  an  arrangement  it  is 
not  strictly  true  that  the  gear  C 
does  not  turn,  yet  its  action  on 
the  gear  D  for  the  interval  of  one 
revolution  of  the  epicyclic  arm 
(that  is,  the  line  joining  the 
centers  of  D  and  C)  is  the  same 
as  though  C  did  not  turn,  since 
the  position  of  C  at  the  end  of 
one  revolution  of  the  arm  is  the 
same  as  at  the  beginning. 


FIG.  201 


Let  it  be  assumed  that  the  gears 
C  and  D  have  the  same  number  of 
teeth. 

Then  the  train  value  =  —  1.     Let  the  arm  db  make  one  turn. 

Required  to  find  the  turns  of  D  and  therefore  of  the  engine  shaft. 


EPICYCLIC  GEAR  TRAINS 


171 


Let  m  represent  the  turns  of  D,  a  the  turns  of  the  arm,  n  the  turns  of  C. 
Solution.     From  equation  (60), 


-1  = 


0-1 


m  -1 

whence  m  =  +2. 

That  is,  the  engine  shaft  will  make  two  turns  every  time  the  gear  C  passes  around  it. 

Example  32.  In  the  three-wheeled  train,  Fig.  203,  let  A  have  55  teeth,  and  C 
have  50.  A  does  not  turn.  To  find  the  turns  of  C  while  the  arm  D  makes  +10 
turns. 

Solution.    Using  the  tabular  method 


Train  locked +10 

Train  unlocked,  arm  fixed -  10 

Resultant  motions .  . 


C 
+  10 

-  10 


Arm 
+  10 
0 


0 

Or  the  wheel  C  turns  —1  while  the  arm  D  turns  +10. 


1 


+  10 


If  the  gear  C  in  Fig.  203  were  given  the  same  number  of  teeth  as  A, 
it  would  not  turn  at  all.  If  there  were  more  teeth  in  C  than  in  A  its 
resultant  number  of  turns  would  be  in  the  same  direction  as  the  arm. 


FIG.  203 


U    U 


FIG.  202 


FIG.  204 


Example  33.  Ferguson's  Paradox.  In  the  device  shown  in  Fig. 
204,  known  as  Ferguson's  Paradox,  all  three  of  the  cases  just  referred 
to  occur  in  one  mechanism.  » 

Let  the  gear  A  have  60  teeth,  C  61  teeth,  E  60  teeth,  F  59  teeth.  B  is  an  idle 
wheel  connecting  each  of  the  others  with  A.  The  arm  D  turns  freely  on  the  axis 
of  A  and  carries  the  axis  which  supports  the  other  gears.  A  is  fixed  to  the  stand 
and  therefore  cannot  turn.  If  the  arm  D  is  given  one  turn  R.H.  (+),  required  to 
find  the  turns  of  C,  E,  and  F. 

Solution.    Using  the  tabular  method 


Train  locked 

Train  unlocked,  arm  fixed , 
Resultant  motions ..... 


A 

D 

C 

E 

F 

+  1 

+  1 

+  1 

+  1 

+  1 

-  1 

0 

—  If 

-  1 

—  f°- 

o      + 1 


o      - 


172 


ELEMENTS  OF  MECHANISM 


Example  34.  Ford  Transmission.  Fig.  205  is  a  diagram  of  the 
planetary  or  epicyclic  transmission  as  used  in  the  Ford  automobile. 
The  shaft  P,  fast  to  the  engine  shaft,  carries  the  piece  L  on  which  are 
clutch  rings.  Between  these  rings  are  other  rings  which  turn  with 
the  drum  K.  The  latter  is  fast  to  the  propeller  shaft  M.  Powerful 
springs  urge  the  clutch  rings  together,  except  when  prevented  by  the 
levers  W.  The  friction  between  the  clutch  rings  serves  to  connect  L  to 


Reverse     Low  Gear   Brake 


Engine 
S&ft 


FIG.  205 

K,  thus  making  a  direct  connection  between  the  engine  shaft  and  the 
propeller  shaft,  furnishing  the  direct  drive  to  the  latter,  which  in  turn 
drives  the  rear  wheels.  This  gives  "full  gear"  forward. 

Fast  to  the  front  side  of  K  is  the  sleeve  S  carrying  the  gear  E,  which 
meshes  with  the  gear  B  running  loosely  on  the  stud  H.  This  stud  is 
carried  by  the  piece  A  which  is  fast  to  the  engine  shaft.  B  is  integral 
with  the  gears  C  and  D.  C  meshes  with  F  on  the  sleeve  N  fast  to  the 
drum  J.  D  meshes  with  G  on  the  hub  of  the  drum  /.  A,  turning  with 
the  engine  and  carrying  the  shaft  H  with  it,  constitutes  the  arm  of  two 
epicyclic  trains.  One  of  these  trains  is  through  the  gears  F,  C,  B,  and 
E.  The  other  train  is  through  the  gears  G,  D,  B,  E. 


EPICYCLIC  GEAR  TRAINS  173 

When  the  left  pedal  in  the  car  is  pressed  forward  a  short  distance 
the  lever  W  is  operated  to  release  the  pressure  of  the  springs  which 
hold  the  clutch  rings  in  contact.  The  engine  then  runs  idly  without 
turning  the  propellor  shaft.  A  further  motion  of  the  left  pedal  applies 
a  brake  band  to  the  drum  J,  holding  it  from  turning,  thus  holding  the 
gear  F  at  rest.  C,  rolling  around  the  stationary  gear  F,  and  B  around 
E,  gives  a  speed  to  E  dependent  upon  the  relative  numbers  of  teeth  in 
F}  C,  B,  and  E. 

This  speed  may  be  calculated  as  follows: 

Using  Eq.  (60)  m  =  turns  of  F  =  0  and  n  the  turns  of  E. 

Assuming  one  turn  of  the  engine,  a  =  +1,  e  =  +fi  X  ||  =  +  T'T- 

7       n  —  1  4  4 

Then  —  =  r r  whence  TO  =  +ry     That  is,  the  gear  E  makes  —  of  a. turn 

for  every  turn  of  the  engine,  in  the  same  direction  as  the  engine,  and  carries  the 
propellor  shaft  with  it. 

If  the  left  pedal  is  allowed  to  come  back  to  its  normal  position,  re- 
leasing the  drum  J,  and  the  hand  lever  operated  to  hold  out  the  clutch, 
then  if  the  middle  pedal  is  pressed  forward,  it  applies  a  brake  to  the  drum 
7,  thus  holding  the  gear  G  stationary.  This  brings  into  action  the 
epicyclic  train  G,  D,  B,  E,  giving  motion  to  E  in  a  direction  opposite 
that  of  the  engine. 

This  may  be  calculated  as  follows: 

Letting  m  represent  the  turns  of  G  and  using  the  same  formula  as  before 

e  =  If  X  H  =  f  • 

5      n  -  1  1 

I  =  Q  _  i*      whence      n  =  —  -• 

That  is  E,  with  the  propellor  shaft,  makes  J  turn  for  each  turn  of  the  engine  and  in 
the  opposite  direction. 

When  the  right  pedal  is  pressed  forward  a  brake  is  applied  to  the 
drum  K,  retarding  or  stopping  the  propellor  shaft.  When  this  is  done 
the  drives  should  of  course  all  be  in  neutral. 

Example  35.  All  Spur  Gear  Differential.  Fig.  206  is  a  diagram  of 
the  arrangement  of  gears  in  an  automobile  differential  where  no  bevel 
gears  are  employed  except  the  pair  which  transmits  the  motion  from 
the  propellor  shaft.  The  large  bevel  gear  A  (usually  a  twisted  bevel) 
is  driven  from  the  propellor  shaft  and  carries  the  cage  B  which  always 
turns  with  A.  This  cage  supports  the  studs  S  on  which  are  the  small 
gears  C  and  D.  There  are  several  pairs  of  these  small  gears  equally 
spaced  around  the  cage  but  one  pair  will  be  sufficient  to  consider  in 
discussing  the  action  of  the  mechanism.  C  meshes  with  the  gear  E 


174 


ELEMENTS  OF   MECHANISM 


which  is  on  the  axle  L,  C  also  meshes  with  D,  as  shown,  and  D  in  turn 
meshes  with  the  gear  F  on  the  axle  R. 

When  the  car  is  moving  in  a  straight  path  so  that  the  rear  wheels 
are  turning  at  the  same  angular  speed,  the  cage  B  and  all  the  gears 
inside  it  revolve  as  a  unit  and  there  is  no  relative  motion  of  the  gears. 
When  the  car  starts  to  turn  a  curve,  the  wheel  which  is  on  the  outside 
of  the  curve  must  travel  farther  than  the  inner  wheel,  and  therefore 


Bearing 


FIG.  206 

must  make  more  turns  in  a  given  time.  The  gears  E,  C,  D,  and  F  then 
begin  to  turn  relative  to  each  other  and  to  the  cage  and  the  action 
becomes  that  of  an  epicyclic  train  with  the  cage,  carrying  the  studs  S, 
as  the  epicyclic  arm. 

This  action  can  be  seen  more  clearly  if  the  car  is  supposed  to  be  standing  still  with 
the  right  wheel  (on  the  axle  R)  jacked  up  clear  of  the  ground  while  the  left  wheel 
rests  on  the  ground  and  is  blocked  to  prevent  it  from  rolling.  Then  E  may  be  con- 
sidered as  the  first  wheel  of  the  epicyclic  train,  C  and  D  intermediate  wheels,  and  F 
the  last  wheel.  The  train  value  between  E  and  F  is  —  1  since  the  gears  E  and  F 
have  the  same  number  of  teeth.  Then,  using  equation  (60)  where  m  =  the  turns 
of  gear  E,  a  =  turns  of  the  cage  and  n  the  turns  of  F. 

m  =0  ,  a  =  +1  (assuming  the  large  bevel  to  make  one  turn). 
n  -  1 


Whence 


-1 


0  -  1 


or  n  =  +2. 


Therefore  with  one  wheel  at  rest  the  other  wheel  will  turn  twice  as  fast  as  the 
large  bevel. 


EPICYCLIC  GEAR  TRAINS 


175 


With  both  wheels  turning,  but  one  of  them  turning  faster  than  the  other,  similar 
relative  motion  takes  place.  For  example,  suppose  that  the  car  is  turning  a  corner 
such  that  the  right  wheel  must  turn  twice  as  fast  as  the  left  one. 

Then,  using  the  same  notation  as  above, 

n  =  2m', 


whence 
or 


ra  —  a 
f  a. 


Example  36.    Triplex  Pulley  Block.     Fig.  207  shows  a  vertical  sec- 
tion and  side  view,  with  part  of  the  casing  removed,  of  a  triplex  pulley 


FIG.  207 


block.  S  is  the  shaft  to  which  the  hand  chain  wheel  A  is  keyed.  Also 
keyed  to  S  is  the  gear  F  meshing  with  the  two  gears  E.  The  gears  E 
turn  on  studs  T  which  are  carried  by  the  arm  B,  the  latter  being  keyed 
to  the  hub  of  the  load  chain  wheel  G.  The  gears  C  are  integral  with  E 
and  mesh  with  the  annular  D  which  is  a  part  of  the  stationary  casing. 
The  mechanism  is  an  epicyclic  train.  F  is  the  first  wheel  of  the  train 
and  has  a  speed  imparted  to  it  by  the  turning  of  the  hand  chain  wheel 
A.  The  annular  D  is  the  last  wheel  of  the  train  and  does  not  turn. 
The  train  value  is 

Teeth  in  F      Teeth  in  C 

Teeth  in  E  X  Teeth  in  D ' 


176 


ELEMENTS  OF  MECHANISM 


FIG.  208 


Assuming  one  turn  of  A,  the  turns  of  the  arm  B  may  be  found,  and, 
therefore,  the  turns  of  G.  Hence,  knowing  the  angular  speed  of  A  and 
its  diameter,  and  the  angular  speed  of  G  and  its  diameter,  the  relative 
linear  speeds  of  the  hand  chain  and  the  load  chain  can  be  calculated. 
The  load  will  then  be  to  the  force  exerted  on  the  hand  chain  as  the 
speed  of  the  hand  chain  is  to  the  speed  of  the  load  chain,  friction  being 
neglected. 

185.  Epicyclic  Bevel  Trains.  Fig.  208  represents  a  common  form 
of  epicyclic  bevel  train,  consisting  of  the  two  bevel-wheels  D  and  E 
attached  to  sleeves  free  to  turn  about  the  shaft  extending  through 

them.  This  shaft  carries  the  cross 
at  F  which  makes  the  bearings  for 
the  idlers  GG  connecting  the  bevels 
D  and  E  (only  one  of  these  idlers 
is  necessary,  although  the  two  are 
used  to  form  a  balanced  pair,  thus 
reducing  friction  and  wear).  The  shaft  F  may  be  given  any  number 
of  turns  by  means  of  the  wheel  A,  at  the  same  time  the  bevel  D  may 
be  turned  as  desired,  and  the  problem  will  be  to  determine  the  resulting 
motion  of  the  bevel  E.  The  shaft  and  cross  F  here  correspond  with 
the  arm  of  the  epicyclic  spur-gear  trains. 

When  the  bevels  are  arranged  in  this  way  the  wheels  D  and  E  must 
have  the  same  number  of  teeth,  and  the  train  value  is  —1.  It  will  be 
found  clearer  in  these  prob- 
lems to  assume  that  the 
motion  is  positive  when  the 
nearer  side  of  the  wheel 
moves  in  a  given  direction, 
say  upward,  in  which  case 
a  downward  motion  would 
be  negative;  or  if  a  down- 
ward motion  is  assumed  as 
positive,  then  upward 
motion  would  be  negative.  j^~\  |/  \ 


M' 


\ 


Example  37.      In    Fig.    209 
B  and  E  are  two  bevel  gears     N 
running  on  shaft  /S,  but  not  fast 
to  it.     Attached  to  the  collar  P, 
which  is  set  screwed  and  keyed  FIG.  209 

to  S,  is  a  stud  T  on  which  turns 

freely  the  gear  D  meshing  with  B  and  E.  B  and  E  are  of  the  same  size  and  T  is  at 
right  angles  with  S.  J  is  a  gear  having  25  teeth  and  driving  the  40-tooth  gear  K 
which  is  fast  to  B.  L  is  a  51-tooth  gear  driven  by  the  17-tooth  gear  H  which  is  fast 


EPICYCLIC   GEAR  TRAINS 


177 


to  E.  N  is  a  45-tooth  gear  fast  to  the  same  shaft  as  J  and  drives  the  20-tooth  gear 
M  which  is  fast  to  S.  It  is  required  to  find  the  speed  of  L  if  J  makes  40  r.p.m. 

Solution.  The  first  step  is  to  pick  out  those  gears  which  are  a  part  of  the  epicyclic 
train.  These  are  evidently  B,  D,  and  E.  The  epicyclic  arm  is  T.  Assume  B  as 
the  first  wheel  of  the  epicyclic  train,  E  the  last  wheel,  and,  letting  m  represent  the 
speed  of  B,  n  the  speed  of  E,  a  the  speed  of  S  and  e  the  train  value  between  E  and 
B.  Also  assume  direction  in  which  /  turns  as  positive.  Using  Eq.  (59), 

e=  -1, 

m  =  —  H  X  40  r.p.m.  =  —25  r.p.m., 
a  =  -fft  X  40  =  -90. 

Then,  substituting  in  Eq.  (59), 

n  =  (-25)  X  (-1)  +  (-90)  -  {(-90)  X(-l)} 
=  25-90-90 
=  —155  r.p.m.  =  speed  of  E. 
Speed  of  L  =  -155  X  (~H)  =  51f. 
Therefore,  L  has  a  speed  of  51  f  r.p.m.  in  the  same  direction  as  J. 

This  problem  may  be  solved  by  the  tabulation  method  also,  the  process  being 
the  same  as  for  an  all  spur  epicyclic  train. 

Example  38.  Bevel  Gear  Differential.  Fig.  210  shows  the  arrange- 
ment of  gears  in  the  differential  of  an  automobile.  Shaft  S  is  driven 
from  the  motor  and  has  keyed  to  it  the  bevel  gear  D  meshing  with  E 
which  turns  loosely  on  the 
hub  of  the  gear  H,  the 
latter  being  keyed  to  the 
axle  of  the  left  wheel. 
E  has  projections  on  it 
which  carry  the  studs  T 
furnishing  bearings  for  the 
gears  R.  There  are  several 
of  these  gears  in  order  to 
distribute  the  load.  The 

gears  R  mesh  with  H  which  ^  ~f\^R 

is,  as  has  been  said,  fast 
to  the  axle  of  the  left 
wheel,  and  with  K  which  is 

fast  to  the  axle  of  the  right  wneel.  When  the  automobile  is  going  straight 
ahead  D  drives  E  and  all  the  other  gears  revolve  as  a  unit  with  E  with- 
out any  relative  motion.  As  soon,  however,  as  the  car  starts  to  turn 
a  corner,  say  toward  the  right,  the  left  wheel  will  have  to  travel  further, 
and  therefore  the  shaft  B  must  turn  faster  than  C.  Then  the  gears 
begin  to  move  relative  to  each  other,  the  action  being  that  of  an  epi- 
cyclic train. 


To  Right  Wheel 


FIG.  210 


178 


ELEMENTS  OF  MECHANISE 


Let  it  be  assumed  that  the  right  wheel  is  jacked  up  so  that  the  axle  C  and  gear 
K  may  turn  freely,  while  the  left  wheel  remains  on  the  ground  and  is  held  from 
turning,  thus  holding  gear  H  from  turning.  Consider  H  as  the  first  wheel  of  the 
train,  E  being  the  arm.  Required  to  find  the  turns  of  C  for  one  turn  (+)  of  E. 

Solution.    Using  equation  (60) 


C 

! 

677. 

\ 

E 

!  j 

D 
\ 

en\ 

it 

~fr 

"-'-' 

_~~~~  r 

/  307. 

\ 

) 

i  ! 

0  -  1 

Whence  n  =  2. 

That  is,  the  right  wheel  will  turn  twice  as  fast  as  the  gear  E. 

Example  39.  Water  Wheel  Governor.     An  epicyclic  bevel  train 
has  been  used  in  connection  with  a  train  containing  a  pair  of  cone  pulleys, 

in  a  form  of  water-wheel 
governor  for  regulating  the 
supply  of  water  to  the 
wheel.  Fig.  211  is  a  dia- 
gram for  this  train,  the 
position  of  the  belt  con- 
necting the  cone  pulleys 
being  regulated  by  a  ball 
governor  connecting  by 
levers  with  the  guiding 
forks  of  the  belt.  The 
governor  is  so  regulated 
that  when  running  at  the 
mean  speed  the  belt  will  be 

in  its  mid-position,  at  which  place  the  turns  of  E  and  D  should  be  equal, 
and  opposite  in  direction,  in  which  case  the  arm  F  will  not  be  turning. 
If  the  belt  moves  up  from  its  mid-position,  and  if  A  turns  as  shown, 
the  arm  F  will  turn  in  the  same  direction  as  the  wheel  E. 

With  the  numbers  of  teeth  as  shown  in  the  figure,  let  it  be  required  to  find  the 

ratio  of  the  diameters  -  if  C  is  to  turn  downward  once  for  25  turns  of  A  in  the  direc- 
x 

tion  shown;  also  to  determine  whether  the  belt  shall  be  crossed  or  open. 

Solution.    Let  E  be  considered  as  the  first  wheel  of  the  train.     Then,  to  use 
equation  (59), 

n  =  turns  A  X  f  $  =  25  X  f$  downward  (+)• 
e  =  -1,  a  =  1. 


u 


or 


Then,  substituting  in  equation  (59), 

25XJ|=  -l(25X 

U.  =      25  X  |?  -  2 


|x|) 


308 
375' 


EPICYCLIC   GEAR  TRAINS 


179 


The  minus  sign  in  this  value  of  -  signifies  that  the  value  m  (in  which  -  first  ap- 

x  x 

pears)  must  be  negative;  that  is,  E  must  turn  in  the  opposite  direction  from  D.    Hence 
the  cone  B  must  turn  in  the  same  direction  as  A  and  the  belt  be  open. 

Example  40.  The  bevel  train  may  be  a  compound  train,  as  shown  in  Fig.  212. 
Here  the'train  value,  instead  of  being  —1,  is  —  *££-  X  ft  —  —*&>  ^  D  *s  considered 
as  the  first  wheel. 


Letting  m  represent  the  turns  of  E,  n  the  turns  of  D,  and  a  the  turns  of  the  arm 
(same  as  of  C)  and  using  Equation  (60),  and  assuming  A  to  make  +  40  turns  and  B 
to  make  —  10  turns, 

_50  =    40  -a 

9   "    -10  -  a  ' 

Whence  a  =  -  ^4<r- 

Or  C  will  turn      0-  times  in  the  same  direction  as  B  and  E. 


CHAPTER  VIII 


INCLINED  PLANE,   WEDGE,    SCREW,   WORM    AND   WHEEL 

186.  Inclined  Plane  and  Wedge.  The  inclined  plane  and  wedge 
will  be  considered  only  as  mechanical  elements  for  producing  motion 
or  exerting  force.  In  this  sense  they  act  essentially  the  same.  In 
Fig.  213,  P  represents  a  wedge,  or  solid,  whose  lower  surface  mn  is 
horizontal,  resting  on  a  horizontal  surface  XX  and  free  to  be  moved 
along  that  surface.  The  upper  surface  mo  is  inclined  at  an  angle  with 
the  horizontal.  In  Fig.  213  the  back  surface  no  is  perpendicular  to  mn. 

S  is  a  slide  which  may  move  up  or  down 
in  the  guides  G,  the  lower  end  being 
inclined  or  beveled  at  the  same  angle  as 
the  upper  surface  of  P,  on  which  it  rests. 
Suppose  that  P  is  moved  to  the  left  a 
distance  rarai,  so  as  to  occupy  the  position 
shown  by  the  dotted  lines.  It  is  evident 
that  S  is  forced  up  a  distance  ddi.  If  the 
length  b  and  height  a  of  P  are  known,  it 
F  010  is  possible  to  calculate  the  amount  S  will 

move  for  any  known  movement  of  P. 

Draw  a  vertical  line  mt  meeting  niiOi  at  t.  Then  mt  =  ddi  since  they 
are  sides  of  a  parallelogram.  The  triangles  niimt  and  wiWiOi  are 
evidently  similar.  Therefore, 


mt 


But 

and 
Therefore 

or 


m\n\ 


=  on, 
mt  =  ddi 

=  mn. 


_  mmi 
mn 

x^  = 

mn 


on 


(61) 


or,  in  words,  the  distance  the  slider  rises  is  equal  to  the  distance  the  wedge 
moves  multiplied  by  the  ratio  of  the  height  of  the  wedge  to  its  length. 

180 


INCLINED  PLANE,   WEDGE,   SCREW,  WORM  AND  WHEEL      181 


In  Fig.  214  a  wedge  is  shown  in  which  the  end  no  is  not  perpendicular 
to  mn.  The  same  method  of  calculating  the  rise  of  the  slider  S  would 
be  used  as  in  the  previous  case  except  that  the  vertical  height  ok  is 
used  in  place  of  the  length  no,  the  shape  of  the  back  end  of  course 
having  no  effect  on  the  motion  of  S. 

The  wedge  in  Fig.  215  is  itself  raised  when  pushed  to  the  left,  due 
to  its  sliding  upon  the  inclined  stationary  surface  of  K,  and  carries  S 


FIG.  214 


FIG.  215 


up  with  it.     It  also  gives  an  additional  rise  to  S  due  to  the  slant  of  the 
surface  mo.     The  resultant  rise  of  S  is,  therefore,  the  sum  of  the  two. 

It  should  be  noticed  that  the  above  laws  hold  true  only  when  the 
direction  of  motion  of  the  slider  S  is  perpendicular  to  the  direction  in 
which  the  wedge  moves. 

187.  Screw  Threads.  If  the  top  surface  mo  of  the  wedge  shown 
in  Fig.  213  is  assumed  to  be  covered  with  a  very  thin  strip  of  flexible 
material  and  this  strip  is  wound  around  a  cylinder  whose  circumference 
is  equal  to  the  length  6,  the  angle  of  _ 
inclination  with  the  horizontal  remaining 
the  same,  it  will  assume  a  helical  form  as 
shown  in  Fig.  216.  If  the  slide  S  has  a 
point  which  reaches  out  and  rests  on  the 
top  surface  of  the  strip,  and  the  cylinder 
is  turned  in  the  direction  of  the  arrow,  S 
will  be  raised.  The  action  of  the  helical 
surface  on  S  is  exactly  the '  same  as  the 
action  of  the  wedge  in  Fig.  213.  One 
complete  turn  of  the  cylinder  will  raise  S 

a  distance  of  P.     One-half  a  turn  will  FIG.  216 

p 
raise  S  a  distance  y>  and  so  on.    In  Fig.  217  a  similar  arrangement  is 

shown,  except  that,  in  this  case,  S  is  stationary  and  the  cylinder  is  free 


182 


ELEMENTS  OF  MECHANISM 


FIG.  217 


to  move  endwise  as  well  as  turn,  the  weight  of  the  cylinder  resting 
on  the  point  of  S  through  the  helical  blade.  Now,  if  the  cylinder 
is  given  one  turn  in  the  same  direction  as  before,  it  will  be  lowered  a 

distance  P. 

A  more  exact  description  of  the  surface 
would  be  to  say  that  it  is  generated  by  a 
radial  line  always  perpendicular  to  the 
axis  of  the  cylinder  and  with  its  inner 
end  in  contact  with  a  helix  of  lead  =  P 
on  the  surface  of  the  cylinder. 

Fig.  218  shows  a  cylinder  with  a  strip 
wound  around  it  in  the  same  way  as  in 
the  preceding  figures  only  the  strip  here 
is  very  much  thicker  and  is  wound 
around  several  times.  In  actually  making 
such  a  cylinder  of  metal  a  solid  cylinder  of  diameter  D  would  be  taken 
and  a  helical  groove  cut  around  it,  the  metal  left  between  the  successive 
turns  of  the  groove  thus  constituting  the  "  helical  strip."  A  cylinder 
so  formed  is  called  a  screw,  the  projecting  part,  which  we  have  called 
the  helical  strip/ 
being  known  as 
the  screw  thread. 
The  action  of- 
such  a  thread  on 
its  follower  is 
exactly  the  same 
as  just  described 
for  Fig.  216  or 
217.  Instead  of 
using^a  single  pro- 
jecting point  or 
surface  for  the 
thread  to  act 


FIG.  218 


FIG.  219 


against,  as  has 

been  assumed   in 

these  figures,  a  hole  with  a  corresponding  thread  inside  it  is  formed, 

this  thread  being  of  the  proper  size,  slant,  and  shape  to  just  fit  into 

the  grooves  of  the  screw.     The  piece  which  contains  such  a  hole  is 

known  as  a  nut.     (See  Fig.  219.) 

188.  Forms  of  Screw  Threads.  There  are  several  forms  of 
threads  in  general  use.  The  more  common  ones  are  shown  in 
Figs.  220  to  223.  In  Fig.  220  is  shown  the  square  thread  used  for 


INCLINED  PLANE,   WEDGE,   SCREW,   WORM   AND  WHEEL      183 


supporting  or  moving  a  load  as  in  a  jack-screw.  In  Fig.  221  is  shown 
the  thread  ordinarily  known  as  the  Acme  thread  which  is  similar  to 
the  square  thread  except  that  its  sides  slope  slightly,  giving  a  stronger 
thread  and  making  it  possible  to  open  and  close  a  split  nut  around  it. 
Such  a  thread  is  used  on  the  lead  screw  of  a  lathe  and  in  similar  places 
where  the  screw  moves  the  carriage  and  where  it  is  necessary  to  sepa- 
rate the  two  halves  of 
the  nut  on  which  it 
acts  when  it  is  desired 
to  break  the  connec- 
tion between  the  screw 
and  the  carriage. 

Figs.  222  and  223 
show  the  V  thread, 
which  is  the  kind 
commonly  used  on 
bolts,  machine  screws, 
and,  in  fact,  for  most 
purposes  where  the  screw  and  nut  serve  for  holding  purposes.  It  is 
also  used  in  light  apparatus  for  causing  motion.  These  two  forms  are 
alike  except  for  a  slight  difference  in  the  angle  of  the  sides  and  a 
difference  in  shape  at  the  point  and  root.  Figs.  224  to  227,  with  the 
accompanying  tables,  show  the  shapes  and  standard  proportions  of  the 
above  forms  of  threads. 

U.   S.   STANDARD  SCREW  THREADS 


P 


B  =  I  V  =  |  P  nearly. 


FIG.  220 


FIG.  221 


FIG.  222        FIG.  223 


Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

1 

20 

3 
4 

10 

If 

6 

3 

3| 

5 

21 

A 

18 

fi 

10 

If 

6| 

31 

3| 

5i 

21 

f 

16 

1 

9 

11 

5 

3£ 

3i 

5f 

2i 

ft 

14 

ft 

9 

11 

5 

3 

3 

5f 

2| 

1 

13 

i 

8 

2 

4* 

4 

3 

6 

2i 

t 

12 
11 

if 
ii 

7 
7 

2f 

2* 

3 

4 

ft 

3 

i! 

H 

11 

.H 

6 

2| 

4 

4f 

2| 

184 


ELEMENTS  OF  MECHANISM 


WHITWORTH  OR  ENGLISH  STANDARD  SCREW  THREAD 


A  =  -L-  =  0.16  P, 


B  =  0.64  P        r  =  0.137  P. 


FIG.  225 


Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

Dia. 

Screw. 

Threads 
per  In. 

} 

20 

f 

11 

1 

8 

If 

5 

3 

31 

18 

H 

11 

i* 

7 

11 

41 

31 

3| 

IT 

16 
14 

1* 

10 
10 

1} 

11 

7 
6 

2 
2i 

42 

3| 

34 

A 

12 
12 

1* 

9 
9 

H 

if 

6 
5 

2| 
2| 

4 

4 

3 

ACME  THREADS 


FIG.  226 


Threads. 

B 

T 

K 

8 

N 

Threads. 

B 

T 

K 

S 

N 

16 

0.048 

0.023 

0.018 

0.039 

0.044 

H 

0.323 

0.232 

0.226 

0.393 

0.399 

10 

0.060 

0.037 

0.032 

0.063 

0.068 

1§ 

0.343 

0.247 

0.242 

0.419 

0.425 

9 

0.066 

0.041 

0.036 

0.070 

0.075 

ITST 

0.354 

0.255 

0.250 

0.433 

0.438 

8 

0.073 

0.046 

0.041 

0.079 

0.084 

11 

0.385 

0.278 

0.273 

0.472 

0.477 

7 

0.081 

0.053 

0.048 

0.090 

0.095 

1* 

0.416 

0.301 

0.296 

0.511 

0.516 

6 

0.093 

0.062 

0.057 

0.105 

0.110 

H 

0.448 

0.324 

0.319 

0.551 

0.556 

5£ 

0.104 

0.070 

0.064 

0.118 

0.123 

l& 

0.479 

0.348 

0.342 

0.590 

0.595 

5 

0.110 

0.074 

0.070 

0.126 

0.131 

i 

0.510 

0.371 

0.366 

0.629 

0.635 

4£ 

0.121 

0.082 

0.077 

0.140 

0.145 

H 

0.541 

0.394 

0.389 

0.669 

0.674 

4 

0.135 

0.093 

0.088 

0.157 

0.163 

1 

0.573 

0.417 

0.412 

0.708 

0.713 

3£ 

0.153 

0.106 

0.101 

0.180 

0.185 

H 

0.604 

0.440 

0.435 

0.747 

0.753 

31 

0.166 

0.116 

0.111 

0.197 

0.202 

I 

0.635 

0.463 

0.458 

0.787 

0.792 

3 

0.177 

0.124 

0.118 

0.210 

0.215 

H 

0.666 

0.487 

0.481 

0.826 

0.831 

2f 

0.198 

0.139 

0.134 

0.236 

0.241 

A 

0.698 

0.510 

0.505 

0.865 

0.870 

2f 

0.210 

0.148 

0.143 

0.252 

0.257 

p 

0.729 

0.533 

0.528 

0.905 

0.910 

2f 

0.229 

0.162 

0.157 

0.275 

0.280 

i 

.0.760 

0.556 

0.551 

0.944 

0.949 

2 

0.260 

0.185 

0.180 

0.315 

0.320 

A 

0.823 

0.603 

0.597 

1.023 

1.028 

U 

0.291 

0.209 

0.203 

0.354 

0.359 

V 

0.885 

0.649 

0.644 

1.101 

1.106 

T\ 

0.948 

0.695 

0.690 

1.180 

1.185 

I 

1.010 

0.741 

0.736 

1.259 

1.264 

INCLINED  PLANE,   WEDGE,   SCREW,   WORM  AND  WHEEL     185 


SQUARE  THREADS 


~£J 


Q-ICM 

"P5** 


FIG.  227 

Threads  per  inch  may  be  about  f  of  the  U.  S.  Standard 
on  both  Acme  and  Square. 

189.  Single  and  Multiple  Threads.  All  of  the  threads  shown  in 
the  preceding  illustrations  are  single  threads;  that  is,  the  threads  are 
formed  by  the  metal  left  between  the  successive  turns  of  a  single  helical 
groove  cut  around  and  around  the  cylinder.  If  two  parallel  helical 
grooves  are  cut,  the  metal  remaining  will  constitute  a  double  thread; 
three  parallel  grooves  will  leave  a  triple  thread,  and  so  on.  The  single, 
double,  and  triple  threads  are  illustrated  in  Figs.  228,  229,  and  230, 
respectively.  It  will  be  noticed  on  the  single  thread  (Fig.  228)  that 


Single  Thread 
FIG.  228 


Double  Thr&ad 
FIG.  229 


Triple  Thread 
FIG.  230 


if  the  ringer  be  placed  on  any  point  of  the  thread,  as  at  A,  and  is  moved 
along  the  thread  until  it  has  gone  once  around  the  screw,  it  will  come 
to  the  point  C.  That  is,  in  moving  once  around  the  screw  the  finger 
has  advanced  along  the  screw  a  distance  AC.  On  the  double  thread 
(Fig.  229)  if  the  finger  starts  at  A  and  follows  the  thread  once  around, 
it  will  come  to  C,  but  this  time  there  is  a  point  D  which  lies  between 
A  and  C.  D  is  the  point  of  the  second  or  parallel  thread.  Similarly,  if 
the  finger  follows  a  thread  in  Fig.  230  once  around  from  A  to  C,  two  points 
D  and  E  will  lie  between  A  and  C.  A  multiple  thread  may  be  used 
when  there  is  need  for  a  fine  thread  having  a  large  "lead"  (see  §  190). 
190.  Lead  and  Pitch  of  a  Screw.  The  distance  AC  which  the 
thread  advances  along  the  screw  in  one  turn  around  is  sometimes 


186 


ELEMENTS  OF  MECHANISM 


called  the  pitch.  A  better  name,  however,  is  the  lead.  This  definition 
of  lead  applies  equally  to  single  and  multiple  threads,  while  the  term 
pitch  is  usually  applied  to  the  distance  from  one  point  to  the  next, 
regardless  of  the  condition  of  the  screw  being  single  or  multiple,  and 
will  be  so  used  in  this  book.  Lead  is  never  used  in  this  sense.  In  the 
case  of  a  single  threaded  screw  the  lead  and  pitch  are  the  same.  If 
a  nut  is  stationary  and  the  screw  is  turned  once  around,  it  will  move 
along  through  the  nut  a  distance  equal  to  the  lead.  If  the  screw  is 
held  from  moving  endwise  but  can  turn,  while  the  nut  is  held  from 
turning  but  is  free  to  move  along  the  screw,  one  turn  of  the  screw  will 
move  the  nut  a  distance  equal  to  the  lead. 

191.  Threads  per  Inch.  The  size  of  a  thread  on  a  screw  is  com- 
monly specified  by  stating  the  number  of  threads  which  the  screw  has 
in  an  inch  of  its  length.  For  example,  Fig.  231  represents  the  side  of  a 


FIG.  231 


FIG.  232 


screw  with  a  scale  laid  against  it  so  that  the  line  M  is  over  the  center 
of  a  groove.  The  line  N,  which  is  an  inch  from  M,  comes  over  the 
center  of  another  groove  or  another  turn  of  the  same  groove  and  there 
are  five  whole  thread  points  between  M  and  N.  This  screw  would  be 
described  as  a  screw  having  five  threads  to  the  inch,  no  account  being 
taken  of  its  being  single  or  double.  In  Fig.  232  the  line  M  is  placed 
over  the  center  of  a  space  and  the  line  N  happens  to  come  over  the  center 
of  a  thread  point  with  seven  whole  thread  points  between.  There  are, 
therefore,  seven  and  one-half  threads  per  inch  on  this  screw.  The  num- 
ber of  threads  per  inch  is  the  reciprocal  of  the  pitch  and  for  a  single 
threaded  screw  is  also  the  reciprocal  of  the  lead. 

192.  Right-hand  and  Left-hand  Threads.  The  thread  may  wind 
around  the  screw  in  such  a  way  that  it  slants  downward  from  right  to 
left  as  one  looks  at  it,  as  shown  in  Fig.  233,  in  which  case  it  is  called  a 
right-hand  thread;  or,  it  may  slant  downward  from  left  to  right  as  one 
looks  at  it,  as  shown  in  Fig.  234,  when  it  is  called  a  left-hand  thread. 
If  the  screw  with  the  right-hand  thread  is  turned  in  the  direction  of 


INCLINED  PLANE,  WEDGE,   SCREW,   WORM  AND  WHEEL      187 


the  Arrow  A,  Fig.  233,  it  will  move  downward  through  the  stationary 
nut,  or  if  the  screw  cannot  move  endwise  the  nut  will  be  drawn  up. 
The  screw  with  the  left-hand  thread  would  have  to  be  turned  in  the 
direction  of  the  arrow  B 
(Fig.  234),  to  move  down- 
ward or  to  draw  the  nut 
up.     If  one  were  looking 
at  the  end  of  a  right- 
hand  screw  and  turned  it 
right-handed  or  clockwise, 
it  would  move   away 
from  him,  whereas  a  left- 
handed  screw  looked  at 
endwise  and  turned  left- 


FIG.  233 


FIG.  234 


handed   or  anti-clockwise 
would  move  away. 

193.  Relation  between  the  Speed  of  a  Screw  or  Nut  and  the  Speed 
of  a  Point  on  the  Wrench  or  Handle.  In  Fig.  235  suppose  the  screw 
S  is  supported  in  a  bearing.  Collars  H  and  B  prevent  it  from  moving 
endwise.  The  lead  of  the  screw  is  P  inches.  S  fits  into  a  nut  N 
which  is  free  to  slide  along  the  guides  G  which  also  keep  it  from  turning. 
A  crank  with  a  handle  K  is  fast  to  the  end  of  the  screw,  the  center  of  K 


FIG.  235 

being  at  a  distance  of  R  inches  from  the  axis  of  the  screw.  It  is  now 
required  to  find  a  method  of  determining  the  relation  between  the 
linear  speed  of  the  handle  K  and  of  the  nut  N.  If  the  crank  is  given 
one  complete  turn  it  will,  of  course,  turn  the  screw  once  and  the  nut 
will  move  along  the  guides  a  distance  P  inches.  While  the  crank  turns 
once  the  center  of  K  moves  over  the  circumference  of  a  circle  whose 
radius  is  R,  therefore  it  moves  over  a  distance  2  irR  inches.  Therefore 

Linear  speed  of  N        P 
Linear  speed  of  K 


(62) 


188 


ELEMENTS  OF   MECHANISM 


Also,  since  the  forces  at  the  two  points  are  inversely  as  the  speeds, 

neglecting  friction, 

Force  at  N  _  2irR 
Force  at  K=     P 

In  Fig.  236,  which  shows  an  ordinary  jack-screw,  the  exact  value  of 
the  speed  ratio  differs  slightly  from  that  expressed  by  Eq.  (62).     Here 

the  point  K  at  which  the  force  is  applied 
rises  with  the  screw  so  that  in  making  a 
complete  turn  the  point  K  moves  over  a 
helix  whose  diameter  is  2  R  and  whose  lead 
is  equal  to  that  of  the  screw.  The  formula 
for  the  length  of  a  helix  is  V  2  irR2  +  P2  so 


that  the  actual  speed  ratio  is 


Linear  speed  of  W      P 

Linear  speed  of  K      \/2^rR2  +  P2 


(64) 


The  lead  (P)  is  so  small  relative  to  R 
that  the   value 


V  2  7r£2  +  P2 


FIG.  236 


differs  only  very  slightly  from  2  irR.  Ac- 
cordingly, although  Eq.  (64)  is  the  correct  one,  Eq.  (62)  is  usually 
accurate  enough  for  all  practical  purposes. 

194.   Compound  or  Differential  Screws.     Fig.  237  illustrates  the 
style  of  screw  known  as  a  differential  screw.     A  part  S  of  the  screw 
itself  has  a  thread  whose  lead  is  P  inches  and  fits  into  a  nut  T  which 
is   a   part   of    the   stationary 
frame.     The  other  end  Si  of 
the    screw    has    a    different 
thread,  of  lead  Pi  inches  which 
fits  the  nut  N.     This  nut  may 
slide  along  the  guides  G  but 
is   held   by  the   guides   from 
turning.     As   the   screw  is 
turned  the  motion  of  the  nut 
is  the  resultant  of  the  move- 
ment of  the  screw  S  through  the  nut  T  and  of  the  nut  N  along  Si.     Sup- 
pose, for  example,  that  P  =  J  in.  and  PI  =  A-  in.,  both  being  right- 
handed  screws.     If  now  the  handle  K  is  turned  right-handed,  as  seen 
from  the  left,  the  whole  screw  moves  along  through  T  toward  the  right 
i  in.  and  if  it  were  not  for  the  thread  Si  N  would  move  to  the  right  j  in. 


FIG.  237 


INCLINED  PLANE,   WEDGE,   SCREW,   WORM  AND  WHEEL     189 


At  the  same  time,  however,  Si  has  drawn  N  back  upon  itself  -£$  in.  so 
that  the  net  movement  of  N  toward  the  right  is  J  in.  —  ^  in.  or  ^  in. 
Again,  suppose  P  =  \  in.  right  hand  and  PI  =  T&  m-  left  hand.  One 
turn  of  the  handle  in  the  same  direction  as  before  will  advance  S  through 
T  |  in.  and  at  the  same  time  carry  N  off  Si  TQ  in.,  so  that  the  net  move- 
ment of  N  to  the  right  is  i  +  re  m-  or  if  m-  A.  device  of  the  first  sort 
may  be  used  for  obtaining  a  very  small  movement  of  the  nut  for  one 
turn  of  the  screw  without  the  necessity  of  using  a  very  fine  thread. 
195.  Examples  on  Velocity  and  Power  of  Screws. 

Example  41.     In  Fig.  238  suppose  it  is  required  to  find  the  load  W,  which,  sus- 
pended from  the  nut  JV,  can  be  raised  by  a  force  of  60  Ib.  applied  at  F.     The  screw 
has  a  lead  of  ^  in.     Assume  that  the  friction  loss  is  40  per 
cent.    Let  R  =  20  in. 

Solution.    While  the  screw  makes  one  turn  F  moves  over 
a  distance  2  IT  20  =  125.66  in.  and  N  rises  |  in. 
Therefore,  F  X  125.66  in.  =  W  X  i  in. 

Since  40  per  cent  is  lost  in  friction  the  net  force  is 

.60  X  60  =  36  Ib. 

Therefore,  36  X  125.66  =  WX  \  in., 

or  W  =  9047.5  Ib. 

The  same  result  would  be  obtained  by  substituting 
directly  in  Eq.  (63). 

Example  42.  In  the  jack-screw  shown  in  Fig.  236,  the 
lead  of  the  screw  is  £  in.  R  =  3  ft.,  6  in.  The  force 
exerted  at  K  is  100  Ib.  To  find  the  weight  W  which  could  FIG.  238 

be  lifted  if  friction  were  neglected. 

Solution.     Equation   (64)  applies  in  this  case  in  finding  the  speed  ratio,  but 
equation  (64)  will  be  very  nearly  correct. 

Speed  of  W  =   ^  in.   =  100 
Speed  of  K  ~  2  TT 42  ~  W' 
Therefore,  W=  2  TT  42  X  100  X  2  =  54,779. 

In  any  case  such  as  this  the  loss  by  friction  would 
be  great  and  would  have  to  be  taken  account  of. 

Example  43.  In  Fig.  239  Pi  =  A  in.  right  hand; 
P2  =  i  in.  right  hand.  To  find  how  many  turns  of 
the  hand  wheel  are  required  to  lower  the  slide  £  in., 
and  to  determine  the  direction  the  wheel  must  be 
turned. 

Solution.  Since  the  outer  screw  is  right  hand 
and  has  a  lead  of  i^  in.  one  turn  of  the  wheel 
right-handed  as  seen  from  above  will  lower  the  outer 
screw  i^  in.  At  the  same  time,  since  the  inner 
screw  is  also  right-handed,  this  one  turn  of  the  wheel 
will  draw  the  inner  screw  into  the  outer  one  |  in.  so  that  the  resultant  downward 
motion  of  the  slide  for  one  turn  of  the  wheel  is  ^  in.  —  £  in.  =  re  in.  Therefore, 
to  lower  it  |  in.  the  wheel  must  be  turned  right-handed  as  seen  from  above  as  many 
times  as  TS  is  contained  in  \  or  8  times. 


FIG.  239 


190  ELEMENTS  OF  MECHANISM 

196.  Rotation  of  Screw  or  Nut  Caused  by  Axial  Pressure.     In  the 
cases  above  considered  the  rotating  force  has  been  assumed  to  act  on 
the  screw  or  nut  in  a  plane  perpendicular  to  the  axis  of  the  screw.     With 
a  screw  of  large  lead  and  relatively  small  diameter,  so  that  the  angle 
which  the  helix  makes  with  the  axis  of  the  screw  is  small,  a  force  acting 
in  the  direction  of  the  axis  may  have  a  component  in  the  direction  to 
cause  rotation  which  is  great  enough  to  overcome  the  frictional  resist- 
ance and  other  resistances  to  turning  and  thus  cause  either  the  screw 
or  the  nut  to  turn.     This  principle  is  made  use  of  iii  small  automatic 
drills  and  screw-drivers,  in  which  axial  pressure  on  the  handle  causes 
the  tool  to  turn.     Such  action  is  not  possible  unless  the  helix  angle  is 
small,  and  the  rotative  component  of  the  force  relatively  large.     It  is 
well  known,  however,  that  constant  jarring  will  cause  nuts  to  work 
loose,  hence  the  necessity  for  cotter  pins  or  double  nuts,  one  serving 
as  a  check  for  the  other. 

197.  Screw  Cutting.     Screws  are  correctly  cut  in  a  lathe  where  the 
cylindrical  blank  is  made  to  rotate  uniformly  on  its  axis,  while  a  tool, 
having  the  same  contour  as  the  space  between  the  threads,  is  made  to 
move  uniformly  on  guides  in  a  path  parallel  to  the  axis  of  the  screwy 
an  amount  equal  to  the  lead  for  each  rotation  of  the  blank.     The  screw 
is  completed  by  successive  cuts,  the  tool  being  advanced  nearer  the  axis 
for  each  cut  until  the  proper  size  is  obtained.    A  nut  can  be  cut  in  the 
same  way  by  using  a  tool  of  the  proper  shape  and  moving  it  away  from 
the  axis  for  successive  cuts. 

Screws  are  also  cut  with  solid  dies  either  by  hand  or  power,  and  with 
proper  dies  and  care  good  work  will  result.  Nuts  are  generally  threaded 
by  means  of  "taps"  which  are  made  of  cylindrical  pieces  of  steel  having 
a  screw-thread  cut  upon  them  of  the  requisite  pitch;  grooves  or  flutes 
are  made  parallel  to  the  axis  to  furnish  cutting  edges,  the  tap  is  tapered 
off  at  the  end  to  allow  it  to  enter  the  nut,  and  the  threads  are  ''backed 
off"  to  supply  the  necessary  clearance.  Before  tapping,  the  nut  must 
have  a  plain  hole  in  it  of  a  diameter  a  little  greater  than  the  root  diam- 
eter of  the  screw  which  it  is  to  fit. 

Screws  cut  by  open  dies  that  are  gradually  closed  in  as  the  screw  is 
being  cut  are  not  accurate,  as  the  screw  is  begun  on  the  outside  of  the 
cylinder  by  the  part  of  the  die  which  must  eventually  cut  the  bottom 
of  the  thread  on  a  considerably  smaller  cylinder.  Thus,  as  the  angle  of 
the  helix  is  greater  the  smaller  the  cylinder,  the  lead  remaining  the  same, 
the  die  at  first  traces  a  groove  having  a  lead  due  to  the  greater  angle 
of  the  helix  at  the  bottom  of  the  thread.  As  the  die-plates  are  made  to 
approach  each  other,  they  tend  to  bring  back  this  helical  groove  to  the 
standard  lead;  this  strains  the  material  of  the  threads,  and  finally  pro- 
duces a  screw  of  a  different  lead  than  that  of  the  die-plates. 


INCLINED  PLANE,   WEDGE,   SCREW,   WORM  AND   WHEEL      191 


198.  Screw  Cutting  in  a  Lathe.  In  cutting  a  screw  thread  in  a 
lathe,  the  stock  on  which  the  thread  is  being  cut  turns  at  a  speed  such 
that  it  will  have  a  surface  speed  suitable  for  the  cutting  tool.  While 
the  work  is  making  one  turn,  the  tool  must  be  fed  along  in  a  direction 


FIG.  240a 

parallel  to  the  axis  of  the  work  a  distance  equal  to  the  lead  of  the  thread 

which  is  being  cut.   Figs.  240a  and  240b  show  one  of  the  simplest  methods 

of     accomplishing    this    result. 

Fig.  240a  is  the  front  view  of  the 

lathe  and  Fig.  240b  the  end  view. 

The  gears  are  lettered  alike  in 

both  views. 

Many  of  the  modern  lathes  use 
a  much  more  elaborate  system 
of  gearing,  but  that  shown  in  the 
figure  serves  to  illustrate  the 
principles  and  is  easier  to  under- 
stand than  the  more  complicated 
ones. 

In  Fig.  240a,  W  is  the  stock  on 
which  the  thread  is  to  be  cut. 


FIG.  240b 


This  is  clamped  to  the  face  plate  by  the  dog  so  that  both  turn  together. 
The  face  plate  is  fast  to  the  spindle  which  is  driven  from  the  cone  pulley 
either  directly  or  through  the  back  gears.  On  the  opposite  end  of  the 


192  ELEMENTS  OF  MECHANISM 

spindle  is  the  gear  A  driving  gear  B  on  the  stud  K  through  one  or  two 
idle  gears  M  and  N  according  to  the  desired  direction  of  rotation.  Fast 
to  the  same  stud,  and,  therefore,  turning  at  the  same  speed  as  B,  is  the 
gear  C.  This  gear  drives  D  through  an  idle  gear.  D  is  fast  to  the 
lead  screw  which  is  embraced  by  a  nut  inside  the  carriage.  The  tool  is 
supported  on  and  moves  along  with  the  carriage. 

Assume  that  the  lead  of  the  thread  to  be  cut  on  the  blank  is  -  part 

1   n 
of  an  inch  and  that  the  lead  of  the  thread  on  the  lead  screw  is  -  part  of 

If 

an  inch.     If  the  blank  makes  a  turns  in  a  unit  of  time,  then  the  distance 

which  the  tool  must  move  in  that  time  must  be  a  X  -;  also  if  6  repre- 

n 

sents  the  number  of  turns  which  the  lead  screw  makes  in  the  same  unit 

of  time,  b  X  -  must  equal  the  distance  the  tool  moves.    Therefore, 
t 

a  x  -=  b  X  T- 

n  t 

1 

Therefore,  5  =  2, 

a      1 

1 

Angular  speed  of  lead  screw  _  lead  of  thread  which  is  being  cut 
Angular  speed  of  blank  lead  of  thread  on  lead  screw 

Now  from  the  laws  governing  wheel  trains 

Angular  speed  of  lead  screw  _  teeth  in  A      teeth  in  C 
Angular  speed  of  blank          teeth  in  B      teeth  in  D ' 
Therefore, 

Teeth  in  A      teeth  in  C  _  lead  of  thread  which  is  being  cut 
Teeth  in  B      teeth  in  D  ~      lead  of  thread  on  lead  screw 

In  any  particular  lathe  the  teeth  in  gears  A  and  B  are  known  quan- 
tities and  cannot  be  changed. 

The  lead  of  the  thread  on  the  lead  screw  is  also  known.  The  gears 
C  and  D  can  be  changed  to  give  the  desired  speed  to  the  lead  screw, 
the  idler  E  being  adjusted  so  as  to  make  proper  connection  between 
them.  If  the  thread  on  the  lead  screw  and  that  being  cut  are  both  right 
hand  or  both  left  hand,  the  lead  screw  must  turn  in  the  same  direction 
as  the  blank.  If  one  thread  is  right  hand  and  the  other  left  hand,  the 
lead  screw  and  the  blank  must  turn  in  opposite  directions.  This  is 
adjusted  by  the  idle  gears  M  and  N. 


INCLINED  PLANE,   WEDGE,   SCREW,   WORM  AND  WHEEL      193 

Example  44.   In  Figs.  240a  and  240b,  assume  that  the  lead  of  the  thread  on  the 
lead  screw  is  f  in.  left  hand;  gear  A  has  20  teeth;  B,  30  teeth;  C,  27  teeth,  and  D, 
54  teeth.     To  find  the  lead  of  the  thread  which  is  being  cut  on  the  blank. 
Solution.     Substituting  in  Eq.  (66), 

20      27  _  lead  of  thread  being  cut 
30X54~  ~T~ 

Solving  this  equation  gives  lead  of  thread  which  is  being  cut  as  |  in.    That  is,  a  screw 
of  8  threads  per  inch  is  being  cut. 

To  determine  whether  a  right-hand  or  a  left-hand  thread  is  being  cut  the  direc- 
tions may  be  followed  through  by  putting  on  arrows.  If  this  were  done  in  the  figure 
the  arrows  would  indicate  that  the  blank  and  the  lead  screw  are  turning  in  the  same 
direction,  therefore,  since  the  lead  screw  has  a  left-hand  thread  the  thread  which  is 
being  cut  is  left  hand.  If  the  lever  R  were  thrown  up  so  as  to  bring  both  idle  gears 
into  use,  the  direction  of  the  lead  screw  would  be  reversed  and  a  right-hand  thread 
would  be  produced. 

Example  45.    Referring  still  to  Figs.  240a  and  240b,  assume  that  the  lead  screw 
and  the  gears  A  and  B  are  the  same  as  in  Example  44.     Let  it  be  required  to  find 
the  number  of  teeth  in  C  and  D  to  cut  20  threads  per  inch  on  the  blank. 
Solution.    Substituting  in  Eq.  (66), 

20      Teeth  in  C  =  ^ 

30      Teeth  in  D  ~  f  " 

Teeth  in  C  _.1V830_1 

LCe'  TeethinD~20X3X20      5* 

Then  any  practical  sized  gears  may  be  used  at  C  and  D  provided  D  has  five  times 
as  many  teeth  as  C;  as,  for  example,  20  teeth  in  D  and  100  teeth  in  C. 

199.  Worm  and  Wheel.*  Fig.  241  is  a  picture  of  a  worm  and 
wheel  mechanism  mounted  on  a  frame  so  as  to  be  used  as  a  model. 
The  worm  is  merely  a  screw  while  the  wheel  is  a  gear  with  teeth  so 
shaped  that  they  mesh  properly  into  the  spaces  of  the  worm  thread. 
The  worm  may  be  right  hand  or  left  hand  and  single  or  multiple 
threaded. 

Just  as  a  screw,  when  turned,  moves  the  nut  along,  so  the  worm,  when 
turned,  pushes  the  teeth  of  the  worm  wheel  along,  causing  the  wheel 
to  turn.  One  turn  of  the  worm  will  move  a  point  on  the  pitch  circle 
of  the  wheel  over  an  arc  equal  in  length  to  the  pitch  or  lead  of  the 
worm.  Therefore,  in  order  to  cause  the  wheel  to  make  a  complete  turn 
the  worm  must  turn  as  many  times  as  the  lead  is  contained  in  the 
circumference  of  the  pitch  circle  of  the  wheel. 

Let  P  represent  the  lead  of  the  worm  and  D  the  pitch  diameter  of 
the  wheel. 

m.  Turns  of  worm      irD  ,.,_» 

Then  = 5 — r— ,  =  —>  (67) 

Turns  of  wheel       P 

p 
or  Turns  of  wheel  =  Turns  of  worm  X  — ~  •  (68) 

*  See  also  Chap.  V. 


194 


ELEMENTS  OF  MECHANISM 


Again,  the  action  of  the  worm  on  the  wheel  may  be  considered  similar 
to  the  action  of  a  rack  on  a  gear.  One  turn  of  a  single-threaded  worm 
is  the  same  as  sliding  a  rack  along  a  distance  equal  to  the  circular  pitch 

of  the  wheel,  or  turning  the  wheel  through  ~  part  of  a  turn,  where  T 
represents  the  number  of  teeth  in  the  wheel.  One  turn  of  a  double- 


Worin 


FIG.  241 

threaded  worm  corresponds  to  moving  a  rack  along  a  distance  equal 

to  twice  the  circular  pitch  of  the  wheel,  or  turning  the  wheel  through 

2 

^  part  of  a  turn. 

Therefore, 

Turns  of  wheel  =  ^  X  turns  of  single-threaded  worm,  (69) 

2 

Turns  of  wheel  =  -^  X  turns  of  double-threaded  worm,  (70) 

3 

Turns  of  wheel  =  ^  X  £wrns  o/  triple-threaded  worm.  (71) 


INCLINED  PLANE,   WEDGE,   SCREW,   WORM   AND  WHEEL      195 


Or,  to  express  the  same  idea  in  more  general  terms, 

Turns  of  wheei  _  Number  of  threads  in  worm 
Turns  of  worm  ~    Number  of  teeth  in  wheel 


(72) 


It  will  be  noticed  that  the  angular  speed  ratio  of  a  worm  and  worm 
wheel  as  expressed  in  Eq.  (72)  is  the  same  as  for  a  pair  of  gears,  consider- 
ing the  worm  as  a  gear.  In  fact  the  worm  and  wheel  do  not  differ 
essentially  from  helical  gears,  with  shafts  at  right  angles,  and  as  has 
been  stated  in  Chapter  V  it  is  not  easy  to  determine  where  the  line  is 
drawn  between  helical  gears  and  worm  and  wheel.  Perhaps  a  general 
method  of  distinguishing  between  the  two  is  as  follows: 

In  helical  gears  each  tooth  on  each  gear  is  only  a  part  of  a  helix  of 
large  lead,  whereas  in  the  worm  and  wheel  the  wheel  teeth  are  short 
lengths  of  helices  of  very  great  lead  while  the  worm  itself  is  a  gear  of 
one,  two  or,  at  any  rate,  very  few  teeth,  each  of  which  winds  around 
once  or  more  times. 

200.    Examples  of  Worm  and  Wheel. 

Example  46.  In  the  worm  and  wheel  mechanism  shown  in  Fig.  242  let  it  be 
required  to  find  the  number  of  turns  of  the  worm  that  would  be  necessary  to  turn 
the  wheel  14  times. 

Solution.    Applying  Eq.  (67), 

Turns  of  worm      ?r  10 


Therefore  Turns  of  worm  =  — ^     =  700  approximately. 


Worm 


'ftch  Circle  of 
Wte&l  10"dfa. 


FIG.  242 


FIG.  243 


Example  47.  The  cylinder  C,  Fig.  243,  is  keyed  to  the  same  shaft  as  the  worm 
wheel.  It  is  required  to  find  the  force  F  which  would  be  necessary  on  the  handle 
in  order  to  raise  the  weight  of  1000  Ib.  if  friction  is  neglected. 


196  ELEMENTS  OF  MECHANISM 

Solution  1.  First  determine  the  ratio  of  the  linear  speeds  of  the  weight  and  the 
point  at  which  the  force  F  is  applied.  If  the  worm  is  assumed  to  make  one  turn'  in 
a  unit  of  time  the  handle  will  have  a  speed  equal  to  the  circumference  of  a  circle 
whose  radius  is  the  distance  from  the  axis  of  the  worm  to  the  center  of  the  handle. 
Therefore,  it  would  be  2  ?r  16  in.  =  32  x  in.  While  the  worm  turns  once  the  wheel 
will  turn 

Lead  .503 , 

Or          — rr;  tlirnS. 


Circumference  of  wheel  pitch  circle  if  12 

Since  the  cylinder  C  turns  at  the  same  angular  speed  as  the  worm  wheel  a  point 

503 
on  its  circumference  will  have  a  linear  speed  of  '—  ^  X  -T  8  =  .3353  nearly. 

Therefore,  since  the  force  is  to  the  weight  as  the  speed  of  the  weight  is  to  the 
speed  of  the  point  at  which  the  force  is  applied, 


Solution  2.  This  problem  might  have  been  solved  by  a  somewhat  shorter  method, 
as  follows:  Assuming  the  worm  single-  threaded,  the  wheel  must  Jhave  as  many 

teeth  as  the  lead  is  contained  in  the  circumference  of  the  pitch  circle,  or  ^^  =  75 

teeth.  Therefore,  one  turn  of  the  worm  will  cause  the  wheel  and  the  cylinder  C  to 
make  ^  of  a  turn  (see  Eq.  69).  Then,  if  the  radius  of  the  crank  were  equal  to  the 
radius  of  C,  the  force  F  would  be  YV  of  the  weight  W.  But  since  the  radius  of  the 
crank  is  4  times  that  of  C  the  force  F  will  be  |  X  A  of  W  or  ^. 

Therefore,  F=  VaT  or  3|  Ibs. 

The  same  result  would  be  obtained  if  the  worm  were  not  single-threaded,  pro- 
vided the  lead  is  the  same.  For  example,  assume  the  worm  double-  threaded,  then 

the  wheel  would  have  2  X  ^7^'  or  150  teeth  and  one  turn  of  the  worm  would  cause 
.oOo 

the  wheel  to  make  Tf  T  of  a  turn  (see  Eq.  70),  which  equals  j$  as  before. 


CHAPTER  IX 


CAMS 

201.  A  cam  is  a  plate,  cylinder,  or  other  solid  having  a  curved  out- 
line or  a  curved  groove,  which  rotates  about  a  fixed  axis  and,  by  its 
rotation,  imparts  motion  to  a  piece  in  contact  with  it,  known  as  the 
follower. 

This  motion  may  be  transmitted  by  sliding  contact ;  but  where  there 
is  much  force  transmitted,  it  is  often  by  accomplished  rolling  contact. 

If  the  action  of  the  piece  is  intermittent,  it  is  sometimes  called  a 
wiper;  that  is,  a  cam,  in  most  places,  is  continuous  in  its  action,  while 
a  wiper  is  always  intermittent : 
but  a  wiper  is  often  called  a 
cam  notwithstanding. 

Fig.  244  is  a  drawing  of  a 
cam  known  as  a  plate  cam,  and 
Fig.  245  a  drawing  of  a  cylinder 
containing  an  irregular  groove 
and  known  as  a  cylindrical 
earn. 

Very  many  machines,  parti- 
cularly automatic  machines, 
depend  largely  upon  cams, 
properly  designed  and  properly 
timed,  to  give  motion  to  the 
various  parts. 

Usually  a  cam  is  designed 
for  the  special  purpose  for 
which  it  is  to  be  used.  In  most 

cases  which  occur  in  practice  FlG  244 

the  condition  to   be   fulfilled 

in  designing  a  cam  does  not  directly  involve  the  speed  ratio,  but  assigns 
a  certain  series  of  definite  positions  which  the  follower  is  to  assume 
while  the  driver  occupies  a  corresponding  series  of  definite  positions. 

The  relations  between  the  successive  positions  of  the  driver  and  fol- 
lower in  a  cam  motion  may  be  represented  by  means  of  a  diagram 
whose  abscissae  are  linear  distances  arbitrarily  chosen  to  represent  angu- 

.  197 


198 


ELEMENTS  OF  MECHANISM 


lar  motion  of  the  cam  and  whose  ordinates  are  the  corresponding  dis- 
placements of  the  follower  from  its  initial  position.  This  is  illustrated 
in  Fig.  246,  where  the  line  Oabc  represents  the  motion  given  by  the  cam. 


FIG.  245 

The  perpendicular  distance  of  any  point  in  the  line  from  the  axis  OY 
represents  the  angular  motion  of  the  driver,  while  the  perpendicular 
distance  of  the  point  from  OX  represents  the  corresponding  movement 
of  the  follower,  from  some  point  considered  as  a  starting  point.  Thus 
the  line  of  motion  Oabc  indicates  that  from  the  position  0  to  4  of  the 
driver,  the  follower  had  no  motion;  from  the  position  4  to  12  of  the 
driver,  the  follower  had  a  uniform  upward  motion  612;  and  from  posi- 


FIG.  246 

tion  12  to  16  of  the  driver,  the  follower  had  a  uniform  downward  motion 
612,  thus  bringing  it  again  to  its  starting-point. 

201A.  Diagrams  for  Cams  giving  Rapid  Movements.  It  is  very 
often  the  case  that  a  cam  is  required  to  give  a  definite  motion  in  a 
short  interval  of  time,  the  nature  of  the  motion  not  being  fixed.  The 
form  of  the  diagram  for  such  a  motion  will  now  be  discussed. 

In  the  diagram  shown  in  connection  with  Fig.  246  the  follower  had 
two  uniform  motions,  and  if  the  cam  be  made  to  revolve  quickly,  quite 
a  shock  will  occur  at  each  of  the  points  where  the  motion  changes,  as 
a,  6,  and  c;  to  obviate  this  the  form  of  the  diagram  can  be  changed, 
provided  it  is  allowable  to  change  the  nature  of  the  motion. 


CAMS 


199 


^ 

s 

s 

x 

^ 

^ 

^ 

?? 

a  
b 

FIG.  246a 


Suppose  a  cam  is  to  raise  a  body  rapidly  from  e  to/  (Fig.  246a),  the 
nature  of  the  motion  to  be  such  that  the  shock  shall  be  as  light  as 
possible. 

For  the  straight  line  Oa  the  case  is  one  of  a  uniform  motion  (as  in  Fig. 
246),  the  body  being  raised  from  e  to  /  in  an  interval  proportional  to 
ob;  here  the  motion  changes  suddenly  at  0  and  a  accompanied  by  a 
perceptible  shock.  The  line  Ocda  would  be  an  improvement,  the  fol- 
lower not  requiring  so  great 
an  impulse  at  the  start  or 
near  the  end  of  the  motion 
each  being  much  more 
gradual  than  before. 

The  body  may  be  made  to 
move  with  a  harmonic 
motion,  the  diagram  for  which 
would  be  drawn  as  follows 
(Fig.  246b): 

Draw  the  semicircle  e5f  on  ef  as  a  diameter;  divide  the  time  line 
Oh  into  a  convenient  number  of  equal  parts  (in  this  case  ten),  and  then 
divide  the  semicircle  into  the  same  number  of  equal  parts;  through  the 
divisions  of  the  semicircle  draw  horizontal  lines  intersecting  the  vertical 
lines  drawn  through  the  corresponding  points  of  division  of  the  time 
line  Oh,  thus  obtain- 
ing points,  as  a,  6,  c, 
etc.  A  smooth  curve 
drawn  through 
these  points  gives 
the  full  curve  Oabcd 
.  .  .  n.  Here  the 
body  or  follower 
receives  a  velocity 

increasing  from  zero  -p, 

at  the   start  to  a 
maximum  at  the  middle  of  its  path,  when  it  is  again  gradually  dimin- 
ished to  zero  at  /,  the  end  of  its  path. 

This  form  of  diagram  gives  very  good  results,  and  is  satisfactory  in 
many  of  its  practical  applications. 

A  body  dropped  from  the  hand  has  no  initial  velocity  at  the  start,  but 
has  a  uniformly  increasing  velocity,  under  the  action  of  gravity,  until  it 
reaches  the  ground;  similarly,  if  the  body  is  thrown  upward  with  the 
velocity  it  had  on  striking  the  ground,  it  will  come  to  rest  at  a  height 
equal  to  that  from  which  it  was  dropped,  and  its  upward  motion  is  the 


200 


ELEMENTS  OF  MECHANISM 


reverse  of  the  downward  one,  that  is,  a  uniformly  retarded  motion. 
(See  §  27.) 

In  designing  a  cam  for  rapid  movement  the  motion  of  the  follower 
should  obey  the  same  law  of  gravity,  and  have  a  uniformly  accelerated 
motion  until  the  middle  of  its  path  is  reached,  then  a  uniformly  retarded 
motion  to  the  end  of  its  path. 

A  body  free  to  fall  descends  through  spaces,  during  successive  units 
of  time,  proportional  to  the  odd  numbers  1,  3,  5,  7,  9,  etc.,  and  the 
total  space  passed  over  equals  the  sum  of  these  spaces. 

To  develop  a  line  of  action  according  to  this  law  upon  the  same  time 
line  Oh,  and  with  the  same  motion  ef,  as  before,  proceed  as  follows: 

Divide  the  time  line  Oh  into  any  even  number  of  equal  parts,  as  ten; 
then  divide  the  line  of  motion  ef  into  successive  spaces  proportional  to 
the  numbers  1,  3,  5,  7,  9,  9,  7,  5,  3,  1,  and  draw  horizontal  lines  through 
the  ends  of  these  spaces,  obtaining  the  intersections  a',  &',  c',  etc.,  with 
the  vertical  lines  through  the  corresponding  time  divisions  1,  2,  3,  etc.; 
a  smooth  curve,  shown  dotted  in  the  figure,  drawn  through  these  points, 
will  give  the  cam  diagram. 

202.  Plate  Cams.  A  plate  cam  imparts  motion  to  a  follower  guided 
so  that  it  is  constrained  to  move  in  a  plane  which  is  perpendicular  to 
the  axis  about  which  the  cam  rotates;  that  is,  in  a 
plane  coincident  with  or  parallel  to  the  plane  in  which 
the  cam  itself  lies.  The  character  of  the  motion  given 
to  the  follower  depends  upon  the  shape  of  the  cam. 
The  follower  may  move  continuously  or  inter- 
mittently; it  may  move  with  uniform  speed  or 
variable  speed;  or  it  may  have  uniform  speed  part 
of  the  time  and  variable  speed  part  of  the  time.  A 
knowledge  of  the  various  types  of  plate  cams,  and  an 
idea  of  the  manner  of  attacking  the  problem  of 
designing  a  cam  for  any  specific  purpose,  can  best  be 
obtained  by  studying  a  number  of  examples. 


Cam  Shaft 


FIG.  247 


Example  48.  A  cam  is  to  be  keyed  to  the  cam  shaft  (Fig. 
247),  which  turns  as  indicated.  The  shape  of  the  cam  is  to  be 
such  that  the  point  of  the  slider  S  will  be  raised  with  uniform 
motion  from  A  to  B  while  the  cam  makes  one-half  a  turn,  and 
lowered  again  to  the  original  position  during  the  second  half- 
turn  of  the  cam.  The  cam  shaft  turns  at  uniform  speed. 

Solution.  (Fig.  248.)  Draw  a  circle  through  A  with  C  as  a  center.  Since  the 
follower  is  to  rise  from  A  to  B  while  the  cam  makes  one-half  a  turn  (or  turns  through 
180°),  and  since  the  cam  shaft  turns  at  uniform  speed,  divide  one-half  of  the  circle 
(AVW)  into  any  number  of  equal  angles  by  the  lines  Ca,  Cb,  Cd,  and  Ce.  Four 
divisions  are  made  in  the  illustration,  although  for  accurate  work  a  greater  number 


CAMS 


201 


would  be  desirable.  The  divisions  are  made  on  the  side  which  is  turning  upward 
toward  the  follower,  that  is,  back  on  the  side/rora'which  the  arrow  is  pointing.  Now, 
divide  the  distance  AB  into  as  many  parts  as  there  are  divisions  in  the  angle  AFW. 
Since  the  follower  is  to  rise  from  A  to  B  with  uniform  motion,  the  divisions  of  AB 
will  be  equal.  That  is,  A  to  1  =  1  to  2  =  2  to  3  =  3  to  B.  When  the  cam  has 


FIG.  248 

made  one-fourth  of  a  half  revolution,  the  line  Ca  will  be  vertical.  A  point  m  on 
this  line,  found  by  swinging  an  arc  through  1  with  center  C,  will  be  the  point  on  the 
cam  which  will  be  at  the  height  Cl  above  the  center  when  the  cam  has  made  one- 
fourth  of  one-half  revolution.  Similarly,  n  will  be  the  point  on  the  cam  which  will 
be  at  2  when  the  cam  has  turned  one-half  of  the  half  revolution,  p  and  r  are  found 
in  the  same  way,  by  drawing  arcs  through  3  and  B  cutting  the  lines  Cd  and  Ce, 
respectively. 


202 


ELEMENTS  OF  MECHANISM 


A  smooth  curve  drawn  through  the  points  A,  m,  n,  p,  and  r  will  be  the  correct 
outline  for  that  portion  of  the  cam  which  will  raise  the  follower  point  from  A  to  B 
as  specified.  Since  the  follower  is  to  be  lowered  from  B  to  A,  also,  with  uniform 
motion  during  the  remaining  half  turn  of  the  cam,  the  other  half  of  the  cam  outline 
will  be  a  duplicate  of  that  already  found. 

Example  49.  Data  the  same  as  for  Example  48,  except  that  the  follower,  instead 
of  having  a  point  shaped  as  in  that  case,  has  a  roller,  as  shown  in  Fig.  249,  on  which 
the  cam  acts.  The  construction 
is  shown  in  Fig.  250.  It  is  neces- 
sary first  to  find  the  outline  of  the 
cam  for  a  follower  like  that  in  Fig. 
248,  the  point  of  the  follower 
being  assumed  to  be  at  the  center 
A  of  the  roller,  Fig.  250.  The 
construction  of  this  curve  is 
exactly  the  same  as  explained  for 
Fig.  248  and  is  lettered  the  same 
in  Fig.  250,  the  curve  itself  being 
drawn  as  a  dot  and  dash  line. 
This  is  called  the  pitch  line  of  the 
cam.  The  next  step  is  to  set  a 
compass  to  a  radius  equal  to  the 
radius  of  the  roller  and,  with 


Cam  Shaft 


FIG.  249 


FIG.  250 


centers  at  frequent  intervals  on  the  pitch  line,  draw  arcs  as  shown  dotted.  The 
true  cam  outline  is  a  smooth  curve  drawn  tangent  to  these  arcs.  It  should  be  noted 
that  the  point  of  tangency  will  not  necessarily  lie  on  the  line  joining  the  center  of 
the  arc  to  the  center  of  the  cam.  For  instance,  consider  the  arc  drawn  with  n  as 
a  center.  The  cam  curve  happens  to  strike  this  arc  at  y,  not  at  the  point  where  the 
arc  cuts  the  line  (76. 


CAMS 


203 


This  condition  often  prevents  the  cam  which  acts  on  a  roller  or  similar  follower 
from  giving  exactly  the  same  motion  as  would  be  obtained  from  the  "pitch  line" 
cam  acting  on  a  pointed  follower.  This  is  likely  to  be  true 
at  convex  places  where  the  motion  changes  suddenly. 

Example  50.  Given  a  follower  with  a  roller  as  shown 
in  Fig.  251.  The  lowest  position  of  the  center  of  the 
roller  is  a  distance  N  above  the  center  of  the  cam  shaft, 
and  the  line  AB  along  which  the  center  of  the  roller  is 
guided  is  a  distance  D  to  the  right  of  a  vertical  line 
through  C.  That  is,  the  center  of  the  cam  shaft  is 
offset  a  distance  D  to  the  left  of  the  line  of  motion  of 
the  center  of  the  follower.  To  draw  the  outline  of  a 
plate  cam  which,  by  turning  as  shown  by  the  arrow, 
shall  raise  the  center  of  the  roller  from  A  to  B  with 
uniform  motion  while  the  cam  makes  one  half  a  turn, 
then  lower  it  again  to  A  during  the  second  half  revolu- 
tion of  the  cam. 

Solution.  Fig.  252  shows  the  solution  of  this  problem. 
Starting  with  C,  locate  the  center  A  by  measuring  a 
distance  D  to  the  right  of  C  and  a  distance  N  above  C.  p 

Draw  a  line  Ck  through  C  and  A.     Since  the  upward 
motion  is  to  take  place  during  one-half  turn  of  the  cam,  measure  back  180°  from  Ck 
and  draw  Ce  (that  is,  kACe  is  a  straight  line).     Divide   the  angle  kCe  into  any 


convenient  number  of  equal  parts  as  before  (in  this  case  four)  by  the  lines  Ca,  Cb, 
Cd.     Divide  AB  into  the  same  number  of  equal  parts,  since  the  follower  is  to  rise 


204 


ELEMENTS  OF  MECHANISM 


with  uniform  speed.  From  C  as  a  center  swing  an  arc  through  1  cutting  Ck  at  5. 
Cut  Ca  with  the  same  arc  at  9.  Make  the  length  9-10  equal  to  5-1.  Then  10  is  one 
point  on  the  pitch  line  of  the  cam.  In  the  same  way  point  12  is  found  by  making  arc 
11-12  equal  to  arc  6-2,  and,  similarly,  all  the  way  around.  The  true  cam  outline  is 
found  as  before  by  drawing  arcs  with  radius  equal  to  the  radii  of  the  roller,  and  with 
centers  on  the  pitch  line,  and  then  drawing  a  smooth  curve  tangent  to  these  arcs. 

Example  61.     Fig.  253  shows  the  method  of  laying  out  a  cam  to  move  a  follower 
from  A  to  B  with  uniform  motion  during  one-quarter  turn  of  the  cam,  hold  it  at  B 


FIG.  253 

during  one-quarter  turn,  lower  it  again  to  A  during  one-quarter  turn,  and  allow  it 
to  remain  at  A  during  the  last  quarter  turn;  the  cam  to  turn  in  the  direction  of  the 
arrow. 

Solution.  Starting  with  C  and  A  as  in  the  preceding  figure,  draw  the  line  kAC. 
It  is  convenient,  for  the  purpose  of  dividing  up  the  angles,  to  draw  a  circle  through 
A  with  C  as  a  center.  With  CAk  as  a  starting  or  reference  line  divide  the  circle  into 
four  equal  parts  by  the  lines  Ce,  Cf  and  Cn.  The  angle  kCe  is  the  angle  through 
which  the  cam  turns  while  the  follower  is  being  raised.  Divide  this  angle  into  equal 
parts,  and  find  points  on  the  cam  pitch  line  as  described  in  Example  50,  the  only 
difference  here  being  that  the  angles  kCa,  kCb,  etc.,  are  smaller.  Point  16  is  the  last 
point  thus  found.  Since  the  follower  is  to  remain  at  rest  at  B  during  the  next  quarter 
turn  of  the  cam,  the  pitch  line  of  the  part  of  the  cam  which  holds  it  up  there  will 
be  an  arc  of  a  circle  drawn  through  16  with  C  as  a  center.  This  part  of  the  cam  will 


CAMS 


205 


extend  from  16  to  18,  the  point  18  being  found  by  extending  the  circle  to  cut  Cf  at 
17  and  making  the  distance  17-18  equal  to  15-16.  The  construction  for  completing 
the  pitch  line  is  exactly  similar,  and  the  cam  curve  proper  is  obtained  as  described 
in  the  previous  examples. 

Example  52.  Fig.  254  is  a  cam  which  raises  the  center  of  the  roller  from  A  to 
B  with  harmonic  motion  during  one-third  of  a  turn,  allows  it  to  drop  to  its  original 
position  instantly,  and  holds  it  there  during  the  remaining  two-thirds  of  a  turn. 
The  angle  kCf,  through  which  the  cam  turns  to  raise  the  roller,  is  laid  off  (120°)  and 
divided  into  an  even  number  of  equal  parts.  Since  the  roller  is  to  rise  with  harmonic 
motion,  a  semicircle  is  drawn  with  AB  as  a  diameter,  and  the  circumference  of  this 
semicircle  is  divided  into  as  many  equal  parts  as  there  are  divisions  in  the  angle  kCf. 


FIG.  254 


From  the  points  of  division  on  this  semicircle  perpendiculars  are  drawn  to  the  line 
AB,  meeting  it  at  points  1,  2,  3.  These  points  are  the  points  of  division  of  AB  to 
be  used  in  finding  the  pitch  line  of  the  cam,  which  is  found  as  previously  described. 
The  last  point  on  the  part  which  raises  the  follower  is  16.  Since  the  follower  is  to 
drop  instantly,  draw  a  straight  line  from  16  to  17,  the  point  where  an  arc  through 
A  cuts  Cf.  The  remainder  of  the  pitch  line  is  a  circle  about  C  through  17  around 
to  A. 

Example  63.  In  Fig.  255  a  cam  is  to  be  placed  on  the  shaft  at  C  to  act  on  a  roller 
centered  at  A  on  the  rocker  ART.  Another  roller  centered  at  T  on  the  rocker  fits 
a  slot  in  the  slider  S.  The  cam  is  to  be  of  such  shape  that,  by  turning  as  shown 
by  the  arrow,  it  will  move  the  slider  with  harmonic  motion  to  the  position  shown 


206 


ELEMENTS  OF  MECHANISM 


FIG.  256 


CAMS 


207 


dotted  during  one-half  turn  of  the  cam  in  the  direction  indicated,  allow  it  to  return 
to  its  original  position,  with  harmonic  motion,  during  the  next  f  of  a  turn  and  allow 
it  to  remain  at  rest  during  the  remaining  |  of  a  turn.  Fig.  256  shows  the  con- 
struction. 

Example  54.  In  Fig.  257  let  it  be  required  to  design  a  cam  to  be  placed  on  shaft 
0  to  raise  slider  A  to  Ai  during  £  of  a  turn  of  the  cam,  allow  it  to  drop  at  once  to  its 
original  position  and  remain  there  during  the  rest  of  the  turn  of  cam,  the  charac- 
ter of  the  motion  of  A  to  be  unimportant  except  that  the  starting  and  stopping 


Cam  Shaft 


FIG.  257 

shall  be  gradual.  The  cam  is  to  act  on  a  roller  on  the  rocker  BCD,  the  rocker  being 
connected  to  the  slider  by  the  link  BA. 

Solution.  Fig.  258.  First  draw  the  motion  diagram  assuming  uniform  motion 
for  A.  This  is  shown  by  the  dotted  line  taa.  Next  substitute  for  this  line  the  line 
shown  full,  the  greater  part  of  which  is  straight,  having  more  slope  than  the  original 
line  and  connected  to  points  t  and  «8  by  curves  drawn  tangent  to  the  sloping  line 
and  tangent  to  horizontal  lines  at  t  and  as. 

Subdivide  into  equal  parts  the  distance  U8,  which  represents  the  |  turn  during 
which  the  motion  of  the  slider  takes  place,  erect  ordinates  at  these  points  cutting 
the  motion  plot  at  points  ai,  02  etc.,  and  project  these  points  on  to  the  path  of  A 
getting  Ait  A*  etc.  From  Ai,  A2,  etc.^with  radius  AB  cut  an  arc  drawn  about  C 
with  radius  CB,  getting  BI,  B2,  etc.  From  these  points  draw  lines  through  C  cutting 
the  arc  of  radius  CD  at  A,  A,  etc.  The  cam  is  found  from  these  points  as  in  previous 
examples. 


208 


ELEMENTS  OF  MECHANISM 


L8 


FIG.  258 


CAMS 


209 


203.  Positive  Motion  Plate  Cams.  It  will  be  noticed  that  in  each 
of  the  cams  which  have  been  discussed,  the  follower  must  be  held  in 
contact  with  the  surface  of  the  cam  by  some  external  force  such  as 
gravity,  or  a  spring.  The  cam  can  only  force  the  follower  away  from 
the  cam  shaft,  while  some  outside  force  must  bring  it  back.  In  case 
it  is  desired  to  make  the  cam  positive  in  its  action  in  either  direction 
without  depending  upon  external  force,  the  cam  must  be  so  constructed 
as  to  act  on  both  sides  of  the  follower's  roller,  or  there  must  be  two 
rollers,  one  on  either  side  of  the  cam.  Fig.  259  shows  a  cam  designed 


FIG.  259 

to  give  the  same  motion  to  the  same  follower  as  in  Fig.  256.  In  Fig. 
259,  however,  the  pitch  line  of  the  cam  is  made  the  center  line  of  a 
groove  of  a  width  equal  to  the  roller  diameter,  thus  enabling  the  cam  to 
move  the  roller  in  either  direction. 

Fig.  260  shows  another  style  of  positive  motion  cam.  The  follower 
consists  of  a  framework  carrying  two  rollers,  one,  roller  (7,  resting  on 
cam  A,  which  is  designed  so  as  to  give  whatever  motion  is  desired  for 
the  follower.  The  other,  roller  D,  rests  on  cam  B,  which  is  designed  to 
be  in  contact  with  roller  D,  the  position  of  the  latter  depending  in  turn 
upon  the  position  of  the  roller  C.  It  would  be  possible  to  have  both 
rollers  touching  the  same  cam,  but  in  that  case  the  movement  of  the 


210 


ELEMENTS   OF  MECHANISM 


follower  could  only  be  chosen  for  one-half  a  turn  of  the  cam,  the  other 
half  being  determined  by  the  shape  of  the  cam  necessary  to  be  in  con- 
tact with  both  rollers. 


FIG.  260 

204.  Plate  Cam  with  Flat^Follower.  —  Example  55.  The  follower 
for  the  cam  shown  in  Fig.  261  has  a  flat  plate  at  its  end  instead  of  a 
roller.  The  cam  is  so  designed  that,  when  it  turns  right-handed,  the 
follower  is  raised  with  harmonic  motion  while  the  cam  makes  one-third 
of  a  turn,  then  remains  at  rest  during  the  next  third  of  a  turn  of  the 
cam  and  is  lowered  with  harmonic  motion  during  the  remaining  third 
of  a  turn. 


CAMS 


211 


If  the  center  line  of  the  guides  in  which  the  follower  moves  does  not 
pass  through  the  center  of  the  cam,  the  shape  of  the  cam  is  not  affected, 
provided  the  direction  is  the  same. 

Solution.  The  method  of  construction  is  as  follows:  Assuming  that  the  follower 
is  shown  in  its  lowest  position,  measure  up  along  a  vertical  line  passing  through  the 
center  of  the  cam  the  distance  08  which  the  follower  is  to  move.  Divide  this  into 
any  even  number  of  harmonic  divisions,  eight  being  used  in  the  drawing.  Lay  back 
the  angle  oEm  equal  to  the  angle  through  which  the  cam  turns  while  the  follower 
is  being  lifted.  Divide  oEm  into  as  many  equal  angles  as  there  are  harmonic  divi- 
sions in  the  line  08.  Through  point  1  swing  an  arc  with  E  as  a  center  cutting  the 
first  radial  line  at  w;  through  w  draw  a  line  perpendicular  to  Ew.  Through  2  draw 


FIG.  261 

an  arc  cutting  the  second  radial  line  at  v  and  draw  through  v  a  line  perpendicular 
to  Ev.  In  a  similar  way  draw  perpendiculars  to  Eu,  Et,  Er,  Ep,  En,  and  Em.  A 
smooth  curve  tangent  to  all  of  these  perpendiculars  will  be  the  outline  of  that  por- 
tion of  the  cam  which  raises  the  follower. 

Since  the  follower  is  to  remain  at  rest  while  the  cam  turns  through  the  next  120° 
the  outline  between  the  line  Em  and  the  line  Ek  120°  away  from  Em  will  be  an  arc 
of  a  circle  through  m  with  E  as  a  center. 

The  outline  of  the  portion  of  the  cam  which  lowers  the  follower  is  found  in  a 
manner  similar  to  that  described  for  raising  it. 

If  the  foot  of  the  follower  made  an  angle  6  with  the  center  line  of  its  path,  0  being 
other  than  90°,  the  construction  lines  at  c,  d,  e,  f,  etc.,  instead  of  being  drawn  per- 
pendicular to  EC,  Ed,  Ee,  Ef,  etc.,  would  be  drawn  making  an  angle  6  with  these  lines. 


212 


ELEMENTS  OF   MECHANISM 


205.  Plate  Cam  with  Flat  Rocker.  —  Example  56.  The  cam  in 
Fig.  262  actuates  the  follower  S  through  the  rocker  R  which  is  pivoted 
at  P.  S  slides  in  guides,  and  remains  still  while  the  cam  makes  a  quar- 
ter turn  right-handed,  then  rises  to  the  upper  dotted  position  with 
harmonic  motion  during  a  quarter  turn  of  the  cam.  During  the  next 


17 


FIG.  262 

'quarter  turn  the  follower  drops  with  harmonic  motion  to  its  original 
position,  and  remains  at  rest  during  the  last  quarter  turn.  The  foot  of 
the  follower  is  a  semicircle  with  center  at  o,  resting  on  the  upper  flat 
surface  of  the  rocker. 

To  find  the  cam  outline,  first  divide  the  distance  06  into  harmonic  spaces,  six 
being  used  in  this  case.  These  points  of  divisions  are  the  successive  positions  of  the 
center  of  the  semicircle.  Draw  arcs  of  the  circle  with  each  of  the  points,  1,  2,  3,  4, 
5,  6,  as  centers.  Draw  the  dotted  circle  K  tangent  to  the  upper  surface  of  the 
rocker  produced.  Next,  draw  the  lines  a,  b,  c,  d,  e,  and  /  tangent  to  circle  K  and 
to  the  arcs  drawn  at  1,  2,  3,  4,  5,  and  6  respectively.  Parallel  to,  and  at  a  distance 
T  from  lines  a,  6,  c,  etc.,  draw  lines  g,  h,  i,  j,  etc.,  cutting  the  vertical  line  through 
the  cam  center  C  at  7,  8,  9,  10,  11,  and  12. 


CAMS 


213 


Since  the  follower  is  to  remain  at  rest  during  a  quarter  turn  of  the  cam,  the  out- 
line of  the  cam  over  the  angle  A  is  an  arc  of  a  circle  with  radius  CE. 

Since  the  upward  movement  takes  place  during  a  quarter  turn,  or  90°,  lay  off 
angle  B  equal  to  90°  and  divide  it  into  as  many  equal  angles  as  there  are  harmonic 
divisions  in  06.  Lay  off  (714  equal  to  (77  and  through  point  14  draw  a  line  making 
the  same  angle  with  (714  that  line  g  makes  with  CE.  Draw  similar  lines  through 
each  of  the  other  radial  lines  C15,  C16,  C17,  CIS,  and  (719.  The  cam  outline  will 
be  a  smooth  curve  tangent  to  all  the  lines  which  have  been  thus  drawn. 

A  similar  construction  is  used  for  finding  the  curve  for  the  part  of  the  cam  which 
lowers  the  follower.  The  last  part  of  the  cam,  over  angle  F,  will  be  a  circular  arc 
to  give  the  period  of  rest. 

206.  Cylindrical  Cams.  —  Example  57.  The  general  appearance  of 
a  cylindrical  cam  has  already  been  shown  (see  Fig.  245.)  Fig.  263  gives 
dimensions  for  the  hub  and  groove  for  a 
cylindrical  cam  which  is  to  hold  a  follower 
still  for  one-eighth  turn  of  the  cam,  move  it 
1  in.  to  the  right  in  a  line  parallel  to  the  axis 
of  the  cam,  with  uniformly  accelerated  and 
uniformly  retarded  motion  (see  §201A) 
while  the  cam  makes  three-eighths  turn, 
hold  it  still  for  one-eighth  turn,  and  return 
it  to  its  original  position  with  similar  motion 
in  three-eighths  turn. 

Solution.  The  solution  of  this  problem  is  shown  in  Fig.  264.  The 
upper  left-hand  view  is  an  end  view  of  the  cam,  the  upper  right-hand 
view  is  a  side  elevation  of  the  cam. 

To  make  the  drawing,  proceed  as  follows: 

Locate  the  center  line  XX'.  On  the  line  XX'  choose  the  point  C  at  any  con- 
venient place  and  draw  the  circle  K  whose  radius  is  equal  to  the  outside  radius  of 
the  cylinder.  Also  draw  the  dotted  circle  P  with  the  radius  equal  to  the  outside 
radius  minus  the  depth  of  the  groove.  Draw  the  vertical  center  line  YY'.  Lay 
back  the  angle  YCB  equal  to  J  of  360°,  that  is,  45°.  This  is  the  angle  through 
which  the  cam  will  turn  before  the  follower  starts  to  move.  Since  the  movement 
of  the  follower  is  to  take  place  during  the  next  three-eighths  of  a  turn,  the  cam  will 
turn  through  the  angle  BCY'  to  give  the  motion  to  the  follower.  Since  the  follower 
is  to  remain  at  rest  during  the  next  one-eighth  turn,  the  angle  Y'CT  equal  to  45° 
will  next  be  drawn,  and  the  remaining  angle  TOY  will  be  the  angle  through  which 
the  cam  will  turn  to  move  the  follower  back  to  its  original  position.  Now,  draw 
the  center  line  MN  at  any  convenient  distance  on  the  right  of  the  figure  already 
drawn,  and  locate  the  point  E  on  this  line  at  a  distance  from  XX'  equal  to  the  out- 
side radius  of  the  cylinder.  On  a  horizontal  line  drawn  through  E  locate  the  points 
F  and  G,  each  at  a  distance  from  E  equal  to  the  radius  of  the  roller  on  which  the 
cam  is  to  act.  Draw  H J  parallel  to  FG  at  a  distance  from  it  equal  to  the  depth  of 
the  groove.  Through  F  and  G  draw  lines  to  the  point  L  where  MN  intersects  the 
axis  XX'.  That  portion  of  the  line  HJ  intersected  between  FL  and  GL  will  be  the 


FIG.  263 


214 


ELEMENTS   OF   MECHANISM 


width  of  the  groove  at  the  bottom.  Before  it  is  possible  to  proceed  further  in  the 
construction  of  this  side  elevation  of  the  cam,  it  is  necessary  to  make  a  development 
of  its  outer  surface.  Draw  the  line  M'N*  equal  in  length  to  the  circumference  of  the 
cylinder. 


Lay  off  M'B'  equal  to  the  length  of  the  arc  YB  and  B'Y'z  equal  to  the  length  of 
the  arc  BY'.  Divide  B'Y'z  into  any  even  number  of  equal  parts,  in  this  case  eight, 
and  letter  points  of  division  a',  &',  c',  d',  e',  /',  and  g'.  Through  the  points  thus 
found  draw  vertical  lines.  On  the  vertical  line  through  M'  lay  off  Af'8  equal  to 


CAMS  215 

the  distance  through  which  the  follower  is  to  move,  and  divide  M'S  into  "gravity" 
divisions  (see  §  201  A),  using  as  many  divisions  as  there  are  equal  divisions  in  B'Y'z- 
Mark  the  points  thus  found  1,  2,  3,  4,  5,  6,  7.  From  1  project  across  to  the  vertical 
through  a'.  From  2  project  to  the  vertical  through  b',  and  so  on,  thus  getting  the 
points  9,  10,  11,  12,  13,  14,  15,  and  16.  A  smooth  curve  drawn  through  these  points 
will  be  the  development  of  the  center  line  of  that  portion  of  the  cam  groove  which 
moves  the  follower  to  the  right.  Make  Y'2T'  equal  to  the  length  of  the  arc  Y'T. 
The  development  of  the  center  line  of  the  groove  between  the  verticals  at  Y'z  and 
Tf  is  a  horizontal  straight  line.  Since  the  return  motion  of  the  follower  is  a  dupli- 
cate of  the  forward  motion,  the  curve  17N',  being  a  duplicate  of  the  curve  B'  16, 
will  be  the  development  of  the  center  line  of  that  portion  of  the  cam  groove  which 
moves  the  follower  back  to  its  original  position. 

The  above  construction  gives  a  development  of  the  center  line  of  the  groove  on 
the  outer  surface  of  the  cylinder.  The  lines  forming  the  development  of  the  sides 
of  the  groove  are  smooth  curves  drawn  tangent  to  arcs,  swung  about  a  series  of 
centers  along  the  line  M'B'  16  17  Nr  with  radii  equal  to  the  radius  of  the  large  end 
of  the  roller  as  shown  in  the  drawing.  Similar  curves  drawn  tangent  to  arcs  swung 
about  the  same  centers  with  a  radius  equal  to  the  radius  of  the  large  end  of  the 
roller  plus  the  thickness  of  the  flange  forming  the  sides  of  the  groove,  will  be  the 
development  of  the  outer  edges  of  these  flanges. 

The  development  of  the  corners  of  the  bottom  of  the  groove  is  constructed  in  the 
same  way,  except  that  the  length  of  the  development  is  less,  because  it  is  a  develop- 
ment of  a  cylinder  of  smaller  radius. 

The  projections  (on  the  side  elevation)  of  the  curves  which  have  just  been  de- 
veloped are  drawn  by  finding  the  projections  corresponding  to  points  r',  s',  t',  v', 
where  these  curves  cut  the  vertical  line,  it  being  borne  in  mind  that  the  vertical  lines 
on  the  development  really  represent  the  developed  positions  of  elements  of  the  cylin- 
der, drawn  through  points  a,  b,  c,  etc.,  which  are  found  by  dividing  the  arcs  BY' 
and  TY  into  divisions  equal  to  the  divisions  in  B'Y'z  and  T'N'.  The  construction 
for  the  points  rf,  s',  t',  and  v'  only  will  be  followed  through  as  the  construction  for  all 
other  points  will  be  exactly  similar.  Through  6  on  the  end  view  draw  an  element 
of  the  cylinder  across  the  side  elevation.  From  e;  where  this  element  intersects 
MN,  lay  off  et  equal  to  b't',  ev  equal  to  b'v',  to  the  right  of  MN  since  t'  and  v'  are  above 
M'N'j  and  es  equal  to  b's'  and  er  equal  to  l'rf,  to  the  left  since  s'  and  r'  are  below 
M'N'.  The  points  r,  s,  t,  v,  are  the  projections  of  points  corresponding  to  r',  s',  tf,  v'. 
Projections  of  all  other  points  where  the  curves  intersect  the  verticals  on  the  develop- 
ment are  found  in  exactly  the  same  way,  and  smooth  curves  drawn  through  the 
points  thus  found  will  be  the  projections  of  the  corners  of  the  groove,  and  of  the 
flange  enclosing  the  groove.  The  projections  of  the  corners  of  the  bottom  of  the 
groove  are  obtained  in  the  same  way  also,  using,  of  course,  elements  through  02,  b2} 
etc.,  instead  of  a  and  b. 

207.  Multiple-turn  Cylindrical  Cam.  Fig.  265  shows  a  cylindrical 
cam  which  requires  two  revolutions  to  complete  the  full  cycle  of  motion 
of  its  follower.  The  method  of  designing  such  a  cam  would  be  similar 
in  principle  to  that  described  for  the  simple  cam  in  Fig.  264.  The  fol- 
lower in  a  case  like  this  may  require  a  special  form  in  order  to  pass 
properly  the  places  where  the  groove  crosses  on  itself.  This  is  sug- 
gested in  Fig.  266.  The  follower  F  is  made  to  fit  the  groove  sidewise, 


216 


ELEMENTS  OF   MECHANISM 


CAMS 


217 


and  is  arranged  to  turn  in  the  sliding  rod,  to  which  it  gives  motion  in  a 
line  parallel  with  the  axis  of  the  cam.  The  guides  for  this  rod  are 
attached  to  the  bearings  of  the  cam,  A  and  B,  which  form  a  part  of  the 
frame  of  the  machine.  A  plan  of  the  follower  is  shown  at  G:  its  elon- 
gated shape  is  necessary  so  that  it  may  properly  cross  the  junctures  of 


the  groove.  In  this  cam  there  is  a  period  of  rest  during  one-half  a  turn  of 
the  cam  at  each  end  of  the  motion;  the  motion  from  one  limit  to  the  other 
is  uniform,  and  consumes  one  and  one-half  uniform  turns  of  the  cam. 

The  cylinder  may  be  increased  in  length,  and  the  groove  may  be  made 
of  any  desirable  lead;  the  period  of  rest  can  be  reduced  to  zero,  or  in- 
creased to  nearly  one  turn  of  the  cam.  A  cylindrical  cam,  having  a 
right-  and  a  left-handed  groove,  is  often  used  to  produce  a  uniform  re- 
ciprocating motion,  the  right-  and  left-handed  threads  or  grooves  passing 
into  each  other  at  the  ends  of  the  motion,  so  that  there  is  no  period  of  rest. 


The  period  of  rest  in  a  cylindrical  cam,  like  that  shown  in  Fig.  266, 
can  be  prolonged  through  nearly  two  turns  of  the  cylinder  by  means 
of  the  device  shown  in  Fig.  267.  A  switch  is  placed  at  the  junction  of 


218  ELEMENTS  OF   MECHANISM 

the  right-  and  left-handed  grooves  with  the  circular  groove,  and  it  is 
provided  that  the  switch  shall  be  capable  of  turning  a  little  in  either 
direction  upon  its  supporting  pin,  while  the  pin  is  capable  of  a  slight 
longitudinal  movement  parallel  with  the  axis  of  the  cylinder.  This 
supporting  pin  is  constantly  urged  to  the  right  by  a  spring,  shown  in  A, 
which  acts  on  a  slide  carrying  the  pin;  when  in  this  position  the  space 
a  between  the  switch  and  the  circular  part  of  the  groove  is  too  small  to 
allow  the  follower  to  pass,  and  when  the  follower  is  in  the  position 
shown  in  Bt  the  spring  is  compressed;  then,  if  the  follower  moves  on, 
the  space  behind  it  is  closed,  as  the  spring  will  tend  to  push  the  sup- 
port to  the  right,  and  swing  the  switch  on  the  follower  as  a  fulcrum. 

If  the  cam  turns  in  the  direction  of  the  arrow,  in  A  the  shuttle- 
shaped  follower  is  entering  the  circular  portion  of  the  groove,  and  leaves 
the  switch  in  a  position  which  will  guide  the  follower  into  the  circular 
groove  when  it  again  reaches  the  switch;  in  B  the  switch  is  pressed 
toward  the  left  to  allow  the  follower  to  pass.  As  motion  continues,  the 
support  of  the  switch  is  pressed  to  the  right,  and  the  switch  is  thrown 
into  the  position  shown  in  C  ready  to  guide  the  shuttle  into  the  return- 
ing groove.  The  period  of  rest  in  this  case  continues  for  about  one 
and  two-thirds  turns  of  the  cylinder. 

Fig.  268  shows  an  arrangement  which  may  be  applied  for  guiding  a 
wire  or  cord  as  it  winds  upon  a  spool.  The  hub  of  the  sheave  is  bored 
to  fit  the  outside  of  the  shaft.  The  shaft  is  stationary  and  has  a  right- 


FIG.  268 


handed  groove  and  a  left-handed  groove  cut  in  it,  and  is  therefore  a 
stationary  cylindrical  cam.  On  the  side  of  the  sheave  is  a  projection 
which  supports  the  pin  on  which  the  specially  constructed  follower  is 
carried.  The  wire  or  cord,  passing  over  the  sheave,  causes  it  to  turn, 
and  as  it  turns  it  receives  a  reciprocating  motion  along  the  axis  of  the 
cam. 


CAMS 


219 


208.  Cylindrical  Cam  Acting  on  a  Lever.  If  the  follower  for  a  cylin- 
drical cam  is  a  pin  or  roller  on  the  end  of  a  lever,  so  that  it  moves  in 
an  arc  instead  of  a  straight  line,  as  in  Fig.  269,  an  exact  construction 
would  require  that  allowance  be  made  for  the  curvature  of  the  path 
when  making  the  development.  This  degree  of  refinement  is  usually 
unnecessary,  from  a  practical  point  of  view,  and  the  cam  may  be  de- 
signed on  the  assumption  that  the  path  of  the  follower  is  a  straight  line 
parallel  to  the  elements  of  the  cylinder. 

If  the  lever  is  in  a  plane  passing  through  the  axis  of  the  cam,  as  in 
Fig.  270,  the  end  of  the  lever  may  be  considered  as  one  tooth  of  a  worm 


Axis_of 

Lever 


FIG.  269 


FIG.  270 


wheel  or  helical  gear,  and  be  given  the  form  of  such  a  tooth.  The  cam 
itself  then  corresponds  to  the  worm  or  to  the  mating  helical  gear  except 
that  its  groove  is  not  necessarily  helical. 

209.  Combinations  of  Two  or  More  Cams.  In  various  automatic 
machines  the  movements  of  parts  which  have  to  be  timed  with  respect 
to  each  other'are  often  obtained  by  the  use  of  two  or  more  cams  properly 
designed,  and  properly  adjusted  to  give  each  piece  its  desired  motion 
at  the  required  time.  Fig.  271  shows  how  a  cylindrical  cam  and  a 
plate  cam  might  be  arranged  to  work  in  combination  with  each  other. 
In  this  case  the  cylindrical  cam  makes  two  revolutions  for  every  one  of 
the  plate  cam.  The  cylinder  R  is  caused  to  swing  back  and  forth  by 
the  lever  S  which,  in  turn,  is  operated  by  the  plate  cam. 

With  the  mechanism  in  the  position  shown,  the  cylindrical  cam  makes 
one-eighth  turn  in  the  direction  shown,  after  which  the  pin  T  starts  to 
move  to  the  right  with  harmonic  motion.  T  moves  to  the  right  the  total 
distance  of  If  in.,  during  three-eighths  of  a  turn  of  the  cylindrical  cam, 
after  which  it  remains  at  rest  for  one-eighth  turn  of  the  cam,  then  re- 


220 


ELEMENTS  OF   MECHANISM 


turns  to  its  original  position  during  the  remaining  three-eighths  turn. 
The  plate  cam  is  so  designed  that,  turning  left-handed  as  shown,  the 
cylinder  R  begins  to  turn  after  T  has  moved  to  the  right  f  of  an  inch. 


FIG.  271 

It  continues  to  turn  with  uniformly  accelerated  and  uniformly  retarded 
motion  until  T  gets  back  again  to  within  f  of  an  inch  of  its  left-hand 
position. 

The  hole  W  will  then  be  in  the  position  now  occupied  by  the  hole  V. 
R  will  then  stop  its  motion  and  T  will  be  inserted  into  the  hole  W. 
During  the  next  revolution  of  the  cylindrical  cam  T  has  a  motion  the 
same  as  before,  and  the  plate  cam  swings  the  cylinder  R  back  to  its 
original  position. 


CHAPTER  X 

FOUR-BAR   LINKAGE.    RELATIVE  VELOCITIES  OF  RIGIDLY 
CONNECTED  POINTS 

210.  The  Four-Bar  Linkage  will  be  discussed  fully  in  a  later  chap- 
ter, the  purpose  of  the  present  chapter  being  to  study  the  relative  linear 
velocities  of  connected  points,  particularly  points  on  a  linkage.  Only 
such  consideration  will  be  given  to  the  linkage  itself  at  this  time  as  to 
make  it  possible  to  study  the  velocities  understandingly. 

Fig.  272  shows  in  a  simple  form  a  mechanism  known  as  a  four-bar 
linkage.  E  is  a  fixed  piece,  such  as  the  frame  of  a  machine.  A 


FIG.  272 

and  D  are  shafts  having  their  bearings  in  E.  The  line  joining  the 
centers  of  A  and  D  is  called  the  line  of  centers.  The  piece  F,  called 
a  crank,  is  keyed  to  A.  H  is  a  similar  crank  keyed  to  D.  The  outer 
ends  of  F  and  H  are  connected  to  each  other  by  the  connecting  rod  K 
and  the  crank  pins  B  and  C.  B  may  be  made  fast  to  K  and  be  free 
to  turn  in  the  hole  in  F  or  it  may  be  fast  to  F  and  free  to  turn  in  K. 
Similarly,  the  pin  C  may  be  free  to  turn  in  either  H  or  K. 

If  shaft  A  is  caused  to  revolve,  the  crank  F  will  revolve  with  it,  the 
center  of  the  pin  B  moving  in  a  circle  whose  center  is  the  center  of  A. 
This  movement  of  B  will,  through  the  connecting  rod  K,  cause  the 

221 


222 


ELEMENTS  OF  MECHANISM 


pin  C  to  move,  and  since  C  can  move  only  in  a  circle  about  the  center 
of  D  the  crank  H  will  be  caused  to  turn,  turning  D  with  it.*     Each 

B J< 0 


FIG.  273 

one  of  the  pieces,  E,  F,  K,  and  H,  is  called  a  link,  and  the  whole 
system  is  called  a  four-bar  linkage. 

It  is  convenient,  in  studying  linkages,  to  indicate  them  by  the  center 
lines  of  the  links,  as  shown  in  Fig.  273  which  represents  the  same  linkage 
as  that  shown  in  Fig.  272. 

211.  Four-Bar  Linkage  with  a  Sliding  Member.  In  Fig.  274,  the 
end  of  the  connecting  rod  carries  a  block,  pivoted  to  it  at  C,  which  slides 


ar 


FIG.  274 

back  and  forth  in  the  circular  slot  as  the  crank  AB  revolves.  The 
center  of  the  slot  is  at  D.  The  center  of  the  crank  pin  C  evidently  has 
the  same  motion  that  it  Would  have  were  it  guided  by  a  crank  of  length 
*  Some  provision  is  necessary  for  passing  the  positions  where  the  connecting  rod 
and  driven  crank  come  into  line  with  each  other. 


FOUR-BAR  LINKAGE 


223 


DC  turning  about  D.  The  mechanism,  therefore,  is  really  a  four-bar 
linkage  with  the  lines  A B  and  DC  as  center  lines  of  the  cranks,  AD  as 
the  line  of  centers,  and  BC  as  the  center  line  of  the  connecting  rod. 

Let  it  now  be  supposed  that  the  slot  is  made  of  greater  radius  than 
that  shown  in  the  figure,  for  example,  with  its  center  at  DI.  Then  the 
equivalent  four-bar  linkage  would  be  ABCDi. 

Carrying  the  same  idea  still  further,  let  the  slot  be  made  straight. 
Then  the  equivalent  center  D  would  be  at  a  point  an  infinite  distance 


FIG.  275 

away.  The  mechanism,  however,  would  still  be  the  equivalent  of  a 
four-bar  linkage,  as  shown  in  Fig.  275,  where  AB  is  one  crank,  the  line 
through  C  perpendicular  to  the  slot  is  the  other  crank,  BC  the  con- 
necting rod,  and  a  line  through  A  parallel  to  the  crank  through  C  is 
the  line  of  centers. 

Fig.  276  shows  the  special  form  in  which  this  linkage  commonly 
occurs,  where  the  center  line  of  the  slot  passes  through  the  center  of 


FIG.  276 

the  shaft  A.  This  is  the  mechanism  formed  by  the  crank  shaft,  crank, 
connecting  rod,  crosshead  and  crosshead  guides  of  the  reciprocating 
steam  engine. 


224  ELEMENTS  OF  MECHANISM 

212.  Relative  Motions  of  the  Links.  In  the  four-bar  linkage  shown 
in  Fig.  277,  A  and  D  are  the  stationary  axes,  AB  and  DC  the  cranks 
and  BC  the  connecting  rod.  If  the  crank  AB  turns  from  the  position 
shown  in  full  lines  to  the  right-hand  dotted  position  —  that  is,  turns 
through  the  angle  BABi  —  the  pin  B  will  travel  over  the  arc  BBi.  This 
will  cause  the  connecting  rod  to  move  and  push  the  pin  C  along  its 
path  to  Ci. 

It  is  apparent  from  the  figure  that  the  length  of  the  arc  CCi  is  not 
equal  to  the  length  of  the  arc  BBi.  In  other  words,  with  the  several 
links  having  the  relative  lengths  as  here  shown,  the  linear  speeds  of 


FIG.  27f 

the  crank  pins  will  differ.  If,  furthermore,  B  is  moved  to  B%,  C  will 
move  to  Cz-  Now  the  arc  BBZ  is  made  equal  to  BB1}  but  CCz  is  evi- 
dently not  equal  to  CCi.  This  construction,  therefore,  suggests  that 
if  the  crank  AB  is  turned  with  uniform  angular  speed  so  that  the  crank 
pin  B  has  a  uniform  linear  speed,  the  crank  pin  C  has  a  varying  linear 
speed  and  the  crank  DC  a  varying  angular  speed.  It  will  be  shown, 
later  (see  §  227),  that  the  motions  of  the  links  of  a  specific  four-bar 
linkage  relative  to  each  other  are  always  the  same  whichever  of  the  four 
links  is  the  stationary  one. 

It  will  be  apparent  also,  when  the  preceding  statement  is  shown  to 
be  true,  that  the  relative  linear  velocities  of  points  on  any  of  the  links  are 
independent  of  the  fixedness  of  the  links. 

213.  Graphical  Representation  of  Motions  and  Linear  Velocities. 
For  convenience  in  graphical  work  in  connection  with  the  study  of 
velocities  it  is  customary  to  represent  the  velocity  of  a  point  by  a 
straight  line  whose  direction  and  length  indicate  the  direction  and 
magnitude  of  the  velocity  of  the  point.  For  example,  let  it  be  assumed 
that  the  block  M  (Fig.  278)  is  sliding  to  the  right  on  the  guide,  at  the 
rate  of  one  foot  per  second.  If  it  is  desired  to  represent  the  velocity  of 
any  point  A  on  this  block  by  a  line,  any  unit  of  length  may  be  assumed 


FOUR-BAR  LINKAGE  225 

to  represent  one  foot  per  second.     Suppose  one  inch  equals  one  foot  per 
second  is  chosen  as  the  convenient  scale;   then  a  line  one  inch  long  is 
drawn  from  A  to  the  right,  parallel  to 
the  guide,  with  an  arrow  head  at  its  end 
pointing  to  the  right. 


If  the  point  is  moving  in  a  curved  Guide 

path  the  line  representing  its  velocity  FIG.  278 

is  drawn  tangent  to  the  path  at  the 
position  of  the   point   on  the  instant  under  consideration. 

If  the  point  has  a  variable  speed  the  length  of  the  line  representing 
its  velocity  is  made  equal  to  the  distance  the  point  would  move  if  it 
continued  for  a  unit  of  time  with  the  same  speed  which  it  has  at  the 
instant  under  consideration. 

214.  Resultant  Motion.     If  a  material  point  receives  a  single  im- 
pulse in  any  direction,  it  will  move  in  that  direction  with  a  certain 
velocity.     If  it  receives  at  the  same  instant  two  impulses  in  different 
directions,  it  will  obey  both,  and  move  in  an  intermediate  direction 
with  a  velocity  differing  from  that  of  either  impulse  alone.     The  posi- 
tion of  the  point  at  the  end  of  the  instant  is  the  same  as  it  would 
have  been  had  the  motions,  due  to  the  impulses,  occurred  in  successive 
instants.     This  would  also  be  true  for  more  than  two  motions.     The 
motion  which  occurs  as  a  consequence  of  two  or  more  impulses  is 
called  the  Resultant,  and  the  separate  motions,  which  the  impulses 
acting  singly  would  have  caused,  are  called  the  Components. 

215.  Parallelogram  of  Motion.     Suppose  the  point  a  (Fig.  279)  to 
have  simultaneously  the  two  component  motions  represented  in  mag- 
nitude and   direction  by  ab  and  ac.     Then  the 
resultant  is  ad,  the  diagonal  of  the  parallelogram 
of  which  the  component  motions  ab  and  ac  are 
the  sides.     Conversely,  the  motion  ad  may  be 
resolved  into  two  components,  one  along  ab,  and 
the  other  along  ac,  by 

drawing  the  parallelogram  abdc,  of  which  it  will 
be  the  diagonal. 

Any  two  component  motions  can  have  but  one 
resultant,  but  a  given  resultant  motion  may 
have  an  infinite  number  of  pairs  of  components. 
In  the  latter  case  there  is  a  definite  solution 
provided  the  directions  of  both  components  are 
known  or  the  magnitude  and  direction  of  one. 
If  the  magnitudes  of  both  components  are  known  there  are  two  possible 
solutions.  Thus  in  Fig.  280,  where  ad  is  the  given  resultant,  if  the 


226  ELEMENTS  OF  MECHANISM 

two  components  have  the  magnitudes  represented  by  ac  and  ab,  the 
directions  ac  and  ab  will  solve  the  problem,  or  the  directions  ac\  and 
obi  will  equally  well  fulfil  the  conditions. 

216.  Parallelepiped  of  Motions.     If  the  three  component  motions 
ab,  ac,  and  ad  (Fig.  281)  are  combined,  their  resultant  of  will  be  the 
diagonal  of  the  parallelepiped  of  which  they  are  the  edges.     The  mo- 
tions ab  and  ac,  being  in  the  same  plane,  can  be  combined  to  form  the 

resultant  ae;  in  the  same  way  ae 
and  ad  can  be  combined,  giving  the 
resultant  af.  Conversely  the  motion 
af  may  be  resolved  into  the  com- 
ponents ab,  ac,  and  ad. 

To    find    the    resultant    of    any 
FIG.  281  -  _. 

number  of  motions :    First,  combine 

any  two  of  them  and  find  their  resultant;  then,  combine  this  resultant 
with  the  third,  thus  obtaining  a  new  resultant,  which  can  be  combined 
with  the  fourth;  and  so  on. 

217.  Composition  and   Resolution  of  Velocities.     If  the  motions 
referred  to  in  the  preceding  paragraphs  are  assumed  to  be  uniform  and 
to  take  place  in  a  unit  of  time,  the  lines  in  Figs.  279,  280,  and  281  may 
be  considered  as  representing  the  velocities  as  well  as  the  motions.     If 
the  motion  of  the  point  is  variable  and  the  lines  ac  and  ab  (Figs.  279 
and  280),  or  ac,  ab,  and  ad  (Fig.  281),  represent  the  velocities  imparted 
to  the  point  a  by  the  impulses  acting  in  the  respective  directions  at 
the  instant,  then  the  lines  ad  (Figs.  279  and  280)  and  af  (Fig.  281)  rep- 
resent the  actual,  or  resultant,  velocities  at  that  particular  instant,  but 
not  the  actual  motions.     In  other  words  (Fig.  279),  ad  indicates  the 
motion  which  a  would  have  in  a  unit  of  time  if  the  impulses  now  acting 
continued  unchanged,  and  therefore  indicates  the  present  velocity,  but 
the  motion  was  assumed  to  be  variable,  therefore  present  conditions 
are  instantaneous  only,  and  ad  does  not  indicate  the  actual  motion  but 
only  the  instantaneous  tendency. 

218.  Relation    between  Linear  Velocities  of  Rigidly   Connected 
Points.     If  two  points  are  so  connected  that  their  distance  apart  is 
invariable  and  if  their  velocities  are  resolved  into  components  at  right 
angles  to  and  along  the  straight  line  connecting  them,  the  components 
along  this  line  of  connection  must  be  equal,  otherwise  the  distance 
between  the  points  would  change. 

This  may  be  seen  by  considering  the  center  line  of  the  connecting 
rod  of  a  four-bar  linkage,  Fig.  282. 

Assuming  that  the  crank  AB  is  turning  at  such  an  angular  speed 
that  the  crank  pin  B  has  a  linear  velocity  represented  by  the  line  Bb, 


FOUR-BAR  LINKAGE  227 

let  it  be  required  to  find  the  line  which  represents,  at  the  same  scale, 
the  linear  velocity  of  C.  Since  Bb  indicates  the  velocity  of  B  at  the 
instant,  B  would,  if  not  restrained  by  the  crank,  move  to  a  point  repre- 
sented by  b  in  a  unit  of  time.  The  same  point  would  be  reached  if 
for  a  part  of  the  unit  of  time  B  should  have  the  velocity  represented 
by  Be,  moving  along  the  line  CB  extended,  to  the  point  e  which  is  the 
foot  of  a  perpendicular  let  fall  from  b  to  CB,  then  from  e  moving  out 
along  this  perpendicular  with  a  velocity  represented  by  eb.  That  is,  the 
velocity  of  B  may  be  considered  as  the  resultant  of  the  component  Be, 
along  the  connecting  rod,  and  the  component  Ee\  (equal  to  eb)  of  rota- 
tion about  some  point  on  the  center  line  of  the  connecting  rod.  Al- 

b^ 


FIG.  282 

though  B  does  not  actually  go  to  6  its  tendency  at  the  instant  is  to  go 
there  in  a  unit  of  time,  and  its  tendency  to  move  along  the  line  CB  is 
such  that,  if  there  were  no  components  at  right  angles  to  CB,  it  would 
go  to  e  in  a  unit  of  time.  It  is  this  tendency  of  B  to  move  along  the  line 
CB,  that  is,  the  component  of  Bb  along  the  line  CB  which  causes  C  to 
move,  and  since  CB  is  a  rigid  rod  which  can  be  neither  lengthened  nor 
shortened,  C  must  have  a  component  velocity  (or  tendency  to  move) 
along  the  line  CB  equal  to  that  of  B,  that  is,  equal  to  Be.  The  direc- 
tion of  the  other  component  which,  when  combined  with  this  one,  will 
give  the  actual  velocity  of  C  must  be  at  right  angles  to  the  rod  BC, 
because  this  component  has  no  effect  along  the  rod. 

The  actual  direction  of  the  velocity  of  C  is  along  a  line  Ck  perpen- 
dicular to  DC.  Then,  to  find  the  velocity  of  C  lay  off  Cf  equal  to  Be 
and  through  /  draw  a  line  perpendicular  to  CB  meeting  Ck  at  h.  Ch 
then  represents  the  velocity  of  C  at  the  same  scale  at  which  the  velocity 
of  B  is  represented.  For  example,  if  Ch  is  found  to  be  three-fourths  as 
long  as  Bb,  it  will  indicate  that  at  that  instant  C  has  a  velocity  which 
is  three-fourths  of  that  of  B.  The  above  method  really  consists,  then, 
of  treating  the  whole  rod  as  if  all  points  were  given  an  impulse  in  the 


228 


ELEMENTS  OF  MECHANISM 


direction  CB  such  as  to  cause  them  to  have  velocities  equal  to  Be  along 
CB  and,  simultaneously,  each  point  were  given  another  impulse  at  right 
angles  to  CB,  causing  all  points  to  turn  about  some  point  on  the  center 
line  of  the  rod. 

219.  Instantaneous  Axis  of  Connecting  Rod.  The  discussion  in 
the  preceding  paragraphs  shows  that  one  end  of  the  connecting  rod  in 
Fig.  283  is  moving  at  the  instant  in  the  direction  Bb,  perpendicular  to 
AB,  while  the  other  end  is  moving  in  the  direction  Ch,  perpendicular 
to  DC.  The  direction  of  B  is  the  same  wherever  the  axis  of  rotation 
is  located  on  the  line  A  B  or  A  B  extended.  Similarly,  the  direction  of 
C  is  the  same  wherever  its  axis  of  rotation  is  located  on  DC  or  DC 


V 

FIG.  283a 


A 
FUed  Axis 


D 

Ftied  Axis 


FIG.  283 


0  B 

FIG.  283b 


extended.  Hence,  both  these  points  B  and  C  might  be  treated  for  the 
instant  as  if  they  were  turning  about  the  point  which  is  common  to  A  B 
and  DC,  that  is,  their  point  of  intersection  0.  B  and  C  are  points  on 
the  connecting  rod  BC,  as  well  as  on  the  cranks,  therefore,  the  whole  rod 
BC  may  be  treated  for  the  instant  as  if  it  were  turning  about  0  as  an 
axis.  The  point  0  is  the  trace  of  a  line  which  is  called  the  instan- 
taneous axis  of  BC.  The  principle  is  the  same  as  if  the  connecting 
rod  were  made  fast  to  a  wheel  which,  for  the  moment,  is  turning  about 
0  as v  an  axis.  It  follows  that  the  velocity  of  C  is  to  the  velocity  of  B 
as  OC  is  to  OB. 

If  the  mechanism  were  such  that  the  motions  of  the  points  B  and  C 
were  not  in  the  same  plane,  the  instantaneous  axis  would  be  found  as 
follows:  Pass  a  plane  through  the  point  B  perpendicular  to  Bb',  the 
motion  Bb  might  then  be  the  result  of  a  revolution  of  B  about  any  axis 
in  that  plane.  In  the  same  manner,  the  motion  of  Ch  might  be  the 
result  of  a  revolution  of  C  about  any  axis  in  the  perpendicular  plane 


FOUR-BAR  LINKAGE  229 

through  C.  The  points  B  and  C,  being  rigidly  connected,  must  rotate 
about  one  axis,  which  in  this  case  will  be  the  intersection  of  the  two 
perpendicular  planes. 

If  the  motions  of  the  two  points  B  and  C  are  in  the  same  plane 
and  parallel,  as  in  Figs.  283a  and  283b,  the  perpendiculars  through 
B  and  C  coincide  and  the  above  method  fails.  Let  Bb  and  Ch  be  the 
velocities  of  the  points  B  and  C  respectively.  To  find  the  instantaneous 
axis  draw  a  right  line  through  the  points  b  and  h  in  each  case  and  note 
the  point  0  where  it  intersects  BC  or  BC  produced.  This  must  be  the 
instantaneous  axis,  for  from  the  similar  triangles  BbO  and  ChO 

Bb  =  OB^ 
Ch~  OC' 

that  is,  the  velocities  of  B  and  C  are  proportional  to  the  distances  of 
these  points  from  0. 

The  trace  of  the  instantaneous  axis  on  the  plane  of  the  drawing  is 
often  called  the  instantaneous  center. 

220.  Centrode.     If  the  position  of  the  instantaneous  axis  of  a 
connecting  rod  or  other  moving  link  is  found  for  a  series  of  positions 
through  the  entire  cycle  of  motion  of  the  linkage,  and  a  smooth  curve 
is  drawn  through  these  different  positions  of  the  instantaneous  axis,  the 
curve  thus  found  is  called  the  centrode  of  the  link  in  question.*    Some 
of  the  properties  of  centrodes  and  the  uses  made  of  them  will  be  taken 
up   later   in   connection   with   certain    special    examples  of   four-bar 
linkages. 

221.  Instantaneous  Axis  and  Centrode  of  a  Rolling  Body.    The 
instantaneous  axis  of  a  rigid  body  which  rolls  without  slipping  upon 
the  surface  of  a  fixed  rigid  body  must  pass  through  all  the  points  in 
which  the  two  bodies  touch  each  other,  for  the  points  in  the  rolling 
body  which  touch  the  fixed  body  at  any  given  instant  must  be  at  rest 
for  the  instant,  and  must,  therefore,  be  in  the  instantaneous  axis.     As 
the  instantaneous  axis  is  a  straight  line,  it  follows  that  rolling  surfaces 
which  touch  each  other  in  more  than  one  point  must  have  all  their 
points  of  contact  in  the  same  straight  line  in  order  that  no  slipping  may 
occur  between  them.     This  property  is  possessed  by  plane,  cylindrical 
and  conical  surfaces  only;  the  terms  cylindrical  and  conical  being  used 
in  a  general  sense,  the  bases  of  the  cylinders  and  cones  having  any  figure 
as  well  as  circles.     The  surface  of  the  fixed  body  is  the  centrode  of  the 
moving  body. 

*  In  reality  the  centrode  is  the  locus  of  the  instantaneous  center,  the  locus  of  the 
instantaneous  axis  being  a  surface  sometimes  called  the  axoid.  The  term  centrode 
is  commonly  used  for  the  locus  of  the  axis  itself  as  well  as  for  the  locus  of  its  trace. 


230 


ELEMENTS  OF   MECHANISM 


Velocities  of  Points  on  Rolling  Bodies.  Let  B,  Fig.  284,  be 
the  center  of  a  wheel  resting  on  a  track  at  the  point  A.  If  a  pull  is 
exerted  at  the  center  along  the  line  Bb,  parallel  to  the  track,  B  will 


move  in  the  direction  Bb.  The  effect  is  the  same  for  the  instant  as  if 
A  B  were  a  crank  turning  about  A  as  an  axis  and  since  AB  is  an  im- 
aginary line  on  the  wheel,  the  whole  wheel  becomes  a  crank  turning 
about  A  as  an  axis. 

Any  other  point  on  the  wheel,  as  E,  must  at  the  instant  in  question 
be  moving  about  A  as  an  axis,  and  therefore  the  direction  of  the  veloc- 
ity of  E  is  Ee  perpendicular  to  the  line  AE.  The  magnitude  of  this 
velocity  is  to  the  magnitude  of  the  velocity  of  B  as  AE  is  to  AB. 

This  velocity  Ee  is  made  up  of  rotation  about  the  center  B  of  the 
wheel  combined  with  a  velocity  parallel  to  the  track  equal  to  the 
velocity  of  B.  That  is,  the  real  velocity  Ee  may  be  considered  as  con- 
sisting of  the  components  Eg  (which  is  equal  and  parallel  to  Bb)  and 
Ef  perpendicular  to  BE  and  of  such  magnitude  that  the  figure  gefE  is  a 
parallelogram. 

223.  Typical  Problems  on  Resolution  and  Composition  of  Veloci- 
ties. The  general  principles  presented  in  the  foregoing  paragraphs 
furnish  the  basis  for  the  determination  of  the  absolute  and  relative 
linear  velocities  of  points  on  a  four-bar  linkage  or,  in  fact,  of  points  on 
any  moving  piece  or  pieces.  A  full  and  clear  understanding  of  the 
methods  of  solving  such  problems  can  be  gained  only  by  studying  the 
actual  solutions  of  a  number  of  cases.  The  following  examples  have 
been  chosen  with  this  end  in  view.  Some  of  the  illustrations  are  taken 
directly  from  actual  machines;  others  are  modified  in  order  to  show 
the  principles  more  clearly;  still  others  are  hypothetical  cases  involving 
constructions  which  are  likely  to  occur  in  actual  cases.  Some  of  the 
examples  involve  special  applications  of  the  four-bar  linkage  which 


FOUR-BAR  LINKAGE 


231 


will  be  analyzed  later  but  which  are  introduced  here  merely  as  examples 
in  velocities. 

Example  68.  In  Fig.  283  let  A  B  and  DC  be  the  center  lines  of  the  cranks  of  a 
four-bar  linkage,  A  and  D  being  the  fixed  axes.  Let  the  linear  velocity  of  the  crank 
pin  B  be  represented  by  the  line  Bb  perpendicular  to  AB.  It  is  required  to  find  the 
lines  which  shall  represent  at  the  same  scale,  the  linear  velocities  of  the  crank  pin 
C  and  of  any  point  E  on  the  center  line  of  the  connecting  rod. 

Solution.  Resolve  the  velocity  Bb  into  the  two  components  Be  and  Br  along 
CB  and  at  right  angles  to  CB,  respectively.  Make  Cf  equal  to  Be  and  through  / 
draw  a  line  perpendicular  to  CB  meeting,  at  h,  a  perpendicular  to  DC,  drawn  through 
C.  Then  Ch  represents  the  velocity  of  C. 

To  find  the  velocity  of  E  it  is  necessary  to  find,  first,  the  direction  in  which  E  is 
moving  at  the  instant.  This  is  done  by  finding  the  instantaneous  axis  of  CB.  AB 
and  DC  are  produced  until  they  meet  at  0,  which  is,  therefore,  the  instantaneous 
axis  of  CB.  OE  is  next  drawn,  and  the  direction  of  the  velocity  E  is  along  a  line 
Em,  perpendicular  to  OE.  The  magnitude  of  this  velocity  is  found  by  laying  off 
En  equal  to  Be  and  drawing  a  line  through  n  perpendicular  to  CB  meeting  Em  at  K. 
EK  is  the  required  line  representing  the  velocity  of  E. 

Example  69.  Fig.  285  shows  in  a  diagrammatic  form  the  crank,  connecting  rod 
and  crosshead  of  a  steam  engine.  Let  Bb  represent  the  velocity  of  B.  It  is  required 


A 
Fixed  Axis 


FIG.  285 


to  find  the  velocity  of  the  center  C  of  the  crosshead  pin  and  of  a  point  E  on  the 
center  line  of  the  connecting  rod. 

Solution.  Bb  is  resolved  into  components  along  and  at  right  angles  to  CB  giving 
Be  as  the  component  along  CB.  Make  Cf  equal  to  Be.  Now  C  is  moving  along  the 
line  NM  parallel  to  the  guides.  Therefore,  the  velocity  of  C  is  Ch,  found  by  drawing 
a  perpendicular  to  CB  at  /  meeting  NM  at  h. 


232 


ELEMENTS  OF  MECHANISM 


The  instantaneous  axis  of  CB  is  at  the  point  0,  where  a  perpendicular  to  NM 
through  C  meets  AB  produced.  Knowing  the  position  of  this  instantaneous  axis  the 
velocity  of  E  can  be  found  as  in  the  preceding  example. 

Example  60.  With  data  the  same  as  for  Example  59,  except  that  the  connecting 
rod  is  a  bent  bar,  as  shown  in  Fig.  286,  let  it  be  required  to  find  the  lines  representing 
the  linear  velocity  of  C  and  E. 

0 


FIG.  286 

Solution.  Since  the  curved  bar  EEC  is  rigid  the  straight  line  EC  may  be 
substituted  for  it  without  affecting  the  relations  of  B  and  C.  Hence  this  line  may 
be  used  as  the  line  of  connection  between  B  and  C  and  the  velocity  of  C  may  be 
found  from  the  known  velocity  of  B  as  in  Example  59.  To  find  the  velocity  of 
E  the  instantaneous  axis  of  BEC  is  found  at  0  as  in  Example  59.  E  is  then  joined 
to  0  and  the  direction  of  the  velocity  of  E  is  along  a  perpendicular  to  OE  through 
E.  Join  C  and  E  and  resolve  the  velocity  Ch  into  components  along  and  at  right 
angles  to  the  line  CE.  The  component  En,  of  the  velocity  of  E,  along  CE  produced 
is  equal  to  the  component  Ck  of  the  velocity  of  C,  along  CE,  the  other  component  in 
each  case  being  at  right  angles  to  CE.  Hence  lay  off  En  equal  to  Ck  and  through 
n  draw  a  line  perpendicular  to  En  meeting  Em  at  m.  Em  then  represents  the 
velocity  of  E. 

Example  61.  In  Fig.  287,  D  is  a  disk  keyed  to  the  shaft  A,  the  center  of  the 
disk  being  at  B.  R  is  a  rod,  one  end  of  which  embraces  the  pin  C  on  the  slider; 
the  other  end  is  enlarged  to  form  the  strap  S  and  contains  a  cylindrical  hole  just 
fitting  over  the  disk  D.  If  the  shaft  A  turns  at  such  an  angular  speed  that  a  linear 
velocity  Bb  is  imparted  to  the  center  B  (Bb  being^perpendicular  to  the  line  from  B 
to  the  center  of  the  shaft  A),  let  it  be  required  totind  the  linear  velocity  imparted 
toC. 


FOUR-BAR  LINKAGE 


233 


Solution.  Since  the  hole  in  S  is  concentric  with  D  its  center  must  coincide  with 
B  in  every  position  which  the  mechanism  occupies  and  the  mechanism  is  the  equiva- 
lent of  the  linkage  in  Fig.  285.  (This  relation  will  be  shown  in  another  way  in  the 
next  chapter.)  Hence  the  problem  becomes  the  same  as  in  Example  59. 

,5 


FIG.  287 

Example  62.  In  Fig.  288  A  B  is  a  crank  turning  about  the  fixed  center  A.  BC 
is  a  connecting  rod  pivoted  to  the  slider  S  at  C  and  prolonged  to  E.  From  E 
another  link  EF  connects  to  the  slider  1 '.  If  Bb  represents  the  velocity  of  B  let  it 
be  required  to  find  the  velocity  of  F  (that  is,  of  the  slider  T). 

Solution.  The  instantaneous  axis  of  BCE  is  at  0,  hence  the  direction  of  the  mo- 
tion of  E  is  perpendicular  to  OE.  Resolve  the  velocity  Bb  into  components  along 


FIG.  288 


and  at  right  angles  to  BCE  and  transfer  the  component  Bb2,  thus  found,  to  Ee2. 
Through  ez  draw  a  perpendicular  to  Ee2  meeting  Ee  at  e.  Ee  is  the  velocity  of  E. 
Next  resolve  Ee  into  components  along  and  perpendicular  to  EF,  getting  Ee3  along 
EF.  Make  Ffr  equal  to  Ee3  and  through  /3  draw  a  perpendicular  to  Ff9  meeting 
the  line  of  motion  of  F  at  /.  Ff  is  the  required  velocity  of  F. 


234  ELEMENTS  OF  MECHANISM 

Example  63.  In  Fig.  289  the  connecting  rod  of  the  four-bar  linkage  is  prolonged 
in  a  curve  to  E.  If  Bbi  represents  the  velocity  of  B  let  it  be  required  to  find  the 
velocity  of  E. 

Solution  No.  1.  The  instantaneous  axis  of  CBE  is  found  at  0,  hence  the  direc- 
tion of  E  is  perpendicular  to  OE.  The  line  of  connection  between  B  and  E  is  the 


-'A*    ^  JD(fjxed 

A  (fixed) 

FIG.  289 

straight  line  BE.  Resolve  the  velocity  of  B  into  components  along  and  perpen- 
dicular to  BE,  getting  Bb2  along  BE.  Make  Ee2  equal  to  Bb2  and  through  e2  draw  a 
perpendicular  to  Eez  meeting  Ee\  at  e\.  Then  Eei  represents  the  velocity  of  E. 

Solution  No.  2.  (Fig.  290.)  Find  the  velocity  of  C  ( =  Cci)  as  in  previous  ex- 
amples. Then  Ccs  is  the  component  of  Cct  which  represents  turning  about  some 
point  on  BC  when  the  component  of  translation  in  direction  CB  is  <7c4  =  Bbt. 
Bba  is  the  component  of  Bb,  which  represents  turning  about  the  same  point  on  BC. 
These  components  of  turning  must  be  proportional  to  the  distances  of  B  and  C  from 
the  axis  about  which  the  turning  occurs.  R  is  this  axis  and  is  found  by  drawing 
63c3  cutting  BC  at  R.  It  should  be  observed  that  R  falls  at  the  foot  of  the  perpen- 
dicular to  BC  let  fall  from  the  instantaneous  axis  of  BC  and  that  its  component  of 
turning  is  zero,  hence  its  true  velocity  is  along  CB  and  is  equal  to  Bb4.  From  the 
above  reasoning  it  follows  that  the  motions  of  all  points  on  the  rod  CBE  relative  ot 
each  other  are  the  same  at  the  instant  as  they  would  be  if  the  rod  were  turning  about 
R  as  a  fixed  axis.  Then  Eez  is  the  component  of  turning  of  E  about  R  and  is  found 

by  making  Eez  perpendicular  to  RE  and  -^  =  TTD  '     Next  make  the  component  of 

translation  Be*  equal  and  parallel  to  .664  and  complete  the  parallelogram  of  which 
Be*  and  Ee3  are  sides.  The  diagonal  Ee\  of  this  parallelogram  represents  the  actual 
velocity  of  E  and  is  the  same  line  as  Ee\  in  Fig.  289.  This  method  of  solution  is  of 


FOUR-BAR  LINKAGE 


235 


value  when  the  instantaneous  axis  falls  off  the  paper  or  in  such  a  position  that  it 
cannot  be  used  conveniently. 


FIG.  290 


Example  64.  (Fig.  291.)  The  long  slider  S  has  attached  to  it,  at  the  pin  C,  the 
link  CB,  the  other  end  of  which  is  connected  to  the  crank  AB.  At  E  is  a  link  EF 
connecting  at^F  with  the  link  DF  which  is  also  connected  to  S.  If  Bbi  represents  the 
linear  velocity  of  B  let  it  be  required  to  find  the  linear  velocity  of  F. 

Solution.  Resolve  Bb\  into  components  along  and  perpendicular  to  CB,  getting 
Bb2.  Make  Cc2  equal  to  Bb2  and  draw  c2ci  perpendicular  to  Cc2  meeting  the  line  of 
motion  of  C  at  d.  Then  Cci  represents  the  velocity  of  C.  The  velocity  of  D  must 
be  the  same  as  that  of  C,  therefore  make  Ddi  equal  to  Cci.  Resolve  Ddi  into  com- 
ponents along  and  perpendicular  to  DF,  getting  Dd2  as  the  component  along  DF. 
Make  Ffa  equal  to  Ddz.  Then  Ffa  is  the  component  of  the  (as  yet  unknown)  veloc- 
.ity  of  F  in  the  direction  DF  when  the  other  component  is1  perpendicular  to  DF. 
Hence  the  line  representing  the  velocity  of  F  must  terminate  somewhere  on  the 
perpendicular  (/2m)  to  F/2  through /2.  (In  other  words  F/2  is  the  projection,  upon  the 
line  DF  produced,  of  the  line  representing  the  velocity  of  F.)  Thus,  by  first  finding 
the  velocity  of  D,  it  is  possible  to  find  the  component  of  translation  along  the  direction 
DF  of  the  point  F  when  the  motion  of  F,  for  the  instant,  is  considered  as  consisting 
of  components  of  translation  along  DF  and  of  turning  about  some  point  on  DF. 

In  a  similar  manner  may  be  found  the  component  Ffa  of  translation  in  the  direc- 
tion FE  when  the  motion  of  F  is  considered  as  consisting  of  translation  along  FE 
and  turning  about  some  point  on  FE.  First  find  the  velocity  of  E  by  locating  the 
instantaneous  axis  0  of  CEB.  The  direction  of  E  is  perpendicular  to  OE  and  the 
length  of  the  line  Eei  representing  its  velocity  is  found  by  making  Ee2  equal  to  Bb2 


236 


ELEMENTS  OF   MECHANISM 


and  drawing  e2ei  perpendicular  to  Ee2,  meeting  Ee\  at  e\.  Next  resolve  Eei  into 
components  along  and  perpendicular  to  FE,  getting  Ee3.  Make  F/3  equal  to  Ee$ 
and  at/s  draw  a  line  perpendicular  to  Ffs.  Since  Ffs  is  the  component  of  the  velocity 
of  F  along  FE  when  the  other  component  is  one  of  turning  about  some  point  on  FE 


(fixed} 


FIG.  291 


the  line  representing  the  velocity  of  F  must  terminate  somewhere  on  the  perpen- 
dicular (/3n)  to  Ffs  through  /3.  It  was  shown  above  that  the  line  representing  the 
velocity  of  F  must  terminate  on  fam.  Therefore  the  end  of  the  desired  line  must  be 
at  the  intersection  of  /2ra  and  fsn,  namely  at  /i,  and  Ffi  represents  the  velocity  of  F. 
Example  65.  In  Fig.  292,  A B  is  a  crank  turning  about  A.  The  crank  pin 
carries  a  block  which  works  in  a  slot  whose  center  line  is  MN.  This  slot  is  in  the 
T-shaped  head  of  the  slider  S  which  moves  in  the  guides  G.  HF  is  the  center  line  of 
the  guides  and  Mn  is  at  right  angles  to  HF.  If  Bb  represents  the  velocity  of  B,  let 
it  be  required  to  find  the  velocity  with  which  S  is  moving  in  the  guides  and  also  the 
rate  at  which  the  block  is  slipping  in  the  slot  MN. 

Solution.  The  component  of  Bb  which  is  parallel  to  HF  will  indicate  the  rate  at 
which  the  block  is  causing  the  slot  to  move,  and  the  component  of  Bb  which  is 
parallel  to  MN  will  indicate  the  rate  at  which  the  block  is  slipping  in  the  slot.  There- 
fore Bb  is  resolved  into  the  two  components  Be  and  Br  (or  eb)  parallel  to  HF  and 

MN,  respectively,  giving  Be  as  the 
velocity  of  S  and  Br  (or  eb)  as  the 
rate  of  slipping  of  the  block  in  the 
slot. 

The  correctness  of  the  above  solu- 
tion canlbe  understood  if  it  beassumed 
that  the  point  B  is  actually  allowed 
to  go  to  b  and  that  instead  of  going 
over  the  path  Bb  it  moves  first  to  e 
along  Be  parallel  to  HF.  In  order  to 
go  there  and  still  remain  in  the  slot 

-p       290  i*  must  move  the  center  line  MN  along 

with  it  until  MN  passes  through  e. 

Then  B  moves  up  along  eb  parallel  to  MN  until  it  reaches  6.     This  latter  motion, 
since  it  is  parallel  to  the  slot,  does  not  move  the  slot  at  all,  but  is  simply  a  motion  of 


Fixed  Axis 


FOUR-BAR  LINKAGE 


237 


slipping  in  the  slot.  Of  course  B  does  not  actually  go  to  b,  since  it  is  constrained  to 
move  in  a  circle  about  A  and  the  next  instant  the  direction  of  the  velocity  of  B  will 
be  different  and  the  rate  of  slipping 
and  the  velocity  of  S  will  have 
changed  accordingly.  As  has  already 
been  pointed  out,  however,  the  velo- 
cities and  directions  shown  in  the 
figure  are  the  ones  which  obtain  for 
the  instant  when  the  mechanism  is  in 
the  position  shown. 

Example  66.  Fig.  293  shows  a 
mechanism  the  same  as  that  shown 
in  Fig.  292,  except  that  the  center 
line  MN  is  not  at  right  angles  to  HF. 
The  method  of  solution  is  the  same  as  FIG.  293 

described  in  the  preceding  example. 

Example  67.  In  Fig.  294,  A  B  is  a  crank  turning  about  the  fixed  axis  A.  The 
crank  pin  whose  center  is  B  carries  a  block  working  in  a  slot  in  the  long  arm  which 
turns  about  the  fixed  axis  G.  The  center  line  MN  of  this  slot  passes  through  G.  If 


Fixed  Axis 


Cran 


Fixed 
Axis  A 


over 


Fixed  Axis  G 


FIG.  294 

Bb  represents  the  velocity  of  B,  let  it  be  required  to  find  the  velocity  of  a  point  H  on 
the  swinging  arm,  on  the  center  line  MN. 

Solution.     The  velocity  of  H  is,  of  course,  the  result  of  an  angular  speed  of  the 
arm  caused  by  the  motion  of  the  block.     This  problem  might  be  solved  by  the 


238 


ELEMENTS  OF  MECHANISM 


principle  of  the  four-bar  linkage  and  will  be  so  analyzed  later.  At  this  point,  how- 
ever, it  will  be  treated  by  means  of  the  principles  of  resolution  of  velocities.  Let  it 
be  assumed  that  the  slot  is  covered  on  the  back  by  a  strip,  as  shown.  Then  a  point 
B\  on  this  strip  which  is  directly  in  line  with  B  is  turning  about  the  center  G.  It  is 
necessary,  first  of  all,  to  find  the  linear  velocity  of  B\.  Bb  is  resolved  into  two  com- 
ponents, Be  and  Br  (or  eb).  Be  is  along  the  direction  in  which  BI  is  moving,  that  is, 
perpendicular  to  BG.  eb  is  parallel  to  the  center  line  of  the  slot,  the  figure  Bebr,  as 
in  the  previous  cases,  being  a  parallelogram.  Be  then  represents  the  velocity  of  B\. 
The  velocity  of  H  is  to  Be  as  GH  is  to  GB.  Since  GB  and  H  are  in  a  straight  line  the 
above  proportion  can  be  made  graphically  by  drawing  from  G  a  straight  line  through 
e  meeting  at  k  the  perpendicular  to  GH  through  H.  The  triangles  GBe  and  GHK 
are  similar,  therefore  HK  is  to  Be  as  GH  is  to  GB.  Hence  HK  is  the  required  veloc- 
ity of  H . 

Example  68.  Fig.  295  shows  a  mechanism  similar  to  that  in  Fig.  294,  except 
that  the  center  line  of  the  slot  does  not  pass  through  G. 

Solution.  The  principles  involved  in  the  solution  are  the  same  as  those  described 
for  Example  67.  The  component  Be  is  perpendicular'to  GB  and  not  to  MN.  H 
k 


Fixed 
Axis 


Fixed 
Axis 


FIG.  295 


FIG.  296 


has  a  velocity  perpendicular  to  GH  and  bearing  the  same  ratio  to  Be  that  GH  bears 
to  GB.  The  similar  triangles  can  be  constructed  by  laying  off  on  GH  a  distance 
GBz  equal  to  GB,  drawing  B$L  perpendicular  to  GBz  and  equal  in  length  to  Be, 
then  drawing  from  G  through  L  to  meet  the  perpendicular  to  GH  through  H  at  K. 
HK  is  then  the  required  velocity  of  H . 


FOUR-BAR  LINKAGE 


239 


Example  69.  The  mechanism  in  Fig.  296  is  the  same  as  that  in  Fig.  295,  except 
that  the  slot  is  curved  instead  of  straight.  The  method  of  solution  is  the  same.  The 
only  new  point  to  be  noticed  is  that  the  line  Br  is  tangent  to  the  center  line  of  the 
slot  at  B 


A  (Fixed} 


FIG.  297 


Example  70.  In  Fig.  297  the  curved  bar  BR  slides  through  the  block  which  is 
pivoted  at  C  on  the  crank  DC.  The  lower  end  of  the  bar  is  carried  by  the  crank 
AB.  C  represents  the  axis  of  the  pin  by  which  the  block  is  attached  to  the  crank 
DC,  while  Cx  represents  the  point  on  the  bar  BR  which  is  inside  the  block  where  the 
axis  of  the  pin  intersects  the  center  line  of  the  bar.  If  Cci  represents  the  linear 


240  ELEMENTS  OF  MECHANISM 

velocity  of  C  and  Bbi  the  velocity  of  the  axis  of  the  pin  B,  let  it  be  required  to  find 
the  linear  velocity  of  a  point  R  on  the  rod. 

Solution  No.  1.  (Fig.  297.)  The  component  of  the  velocity  of  Cx  normal  to 
the  sliding  surfaces,  when  paired  with  another  component  in  the  direction  of  sliding, 
must  be  equal  to  the  component  of  Cci  normal  to  the  slide  when  paired  with  another 
component  along  the  slide.  This  component  normal  to  the  slide  is  CxCs.  Also, 
the  component  CxCz  along  the  line  joining  Cx  and  B,  when  paired  with  another  com- 
ponent perpendicular  to  CzC2,  is  equal  to  Bb*.  Hence  the  velocity  of  Cx  is  Czc4  and 
the  instantaneous  axis  of  BCXR  is  at  0.  The  direction  of  R  is  therefore  perpen- 
dicular to  OR  and  the  length  of  the  line  Rn  which  represents  the  velocity  of  #  is 

found  from  the  equation   77^-  =  ^77-- 


Solution  No.  2.  (Fig.  298.)  First  assume  that  the  axis  c  of  the  block  is  fixed 
(that  is,  Cci  =  0).  Then  the  instantaneous  axis  of  BR  would  be  at  Oi  and  the  line 
representing  the  velocity  of  R  would  be  represented  by  Rr6  perpendicular  to  OiR 

and  of  a  length  such  that  §£  =  &| 
£>DI      U\r> 

Next  assume  B  to  be  fixed  (that  is,  Bbi  =  0).  Then  the  velocity  of  Cx  would 
be  represented  by  CaA,  found  as  in  Example  68,  and  the  velocity  of  R  would  be 

E>«  75  7~> 

represented  by  Rr6  perpendicular  to  BR  and  of  a  length  such  that  77-^  =  -577-  . 

Cxc6          -OUa; 

The  line  representing  the  actual  velocity  of  R  is  Rn  which  is  the  diagonal  of  the 
parallelogram  of  which  Rr6  and  Rr&  are  sides. 

Example  71.  In  the  mechanism  shown  in  Fig.  299,  a  wheel  turning  about  the 
axis  D  has  pinned  to  it  at  C  a  link  connected  to  the  end  of  the  crank  AB.  The 
wheel  also  carries  a  pin  E  on  which  is  a  block.  Through  the  block  slides  the  rod 
RHF.  The  point  H  is  connected  to  the  pin  B  and  the  end  F  is  attached  to  the 
slider.  The  wheel  is  turning  at  such  an  angular  speed  that  the  axes  of  the  pins 
C  and  E  have  velocities  represented  by  Cci  and  Eei  respectively.  Let  it  be  required 
to  find  the  line  representing  the  velocity  of  any  point  as  R  on  the  rod  RHF. 

Solution.  First  find  the  velocity  of  B  and  resolve  this  velocity  Bbi  into  com- 
ponents along  and  perpendicular  to  BH,  thus  getting  Bbs.  The  instantaneous  axis 
of  RHF  lies  somewhere  on  the  perpendicular  to  XX  through  F  but  its  exact  position 
on  this  line  is  not  yet  known.  Hence  the  velocity  of  H  cannot  be  found  as  yet, 
H  is  a  point  which  is  common  to  BH  and  RHF  and  the  component  of  its  velocity 
along  BH  must  be  equal  to  Z?63.  Therefore  the  component  along  BH  of  the  velocity 
of  any  point  on  RHF  which  lies  along  BH  or  BH  produced  must  also  equal  Bb. 
(it  being  understood  that  the  other  component  is  perpendicular  to  BH).  If  RHF 
is  assumed  to  be  enlarged  to  include  the  point  K  where  BH  produced  intersects  the 
perpendicular  to  XX  through  F,  then  K,  being  a  point  on  RHF  lying  along  BH<s 
will  have  the  component  Kk3  equal  to  Bb3.  The  actual  velocity  of  K  must  be  per- 
pendicular to  FK,  hence  Kki  is  the  velocity  of  K.  Let  Ex  represent  the  point  on 
the  axis  of  RHF  where  it  intersects  the  axis  of  the  pin  E.  Then  resolve  Kki  into 
components  perpendicular  to  and  along  KEX  getting  Kk4.  Make  Exe*  equal  to 
Kki  and  draw  a  perpendicular  to  Exe\  through  e4.  Next  resolve  the  velocity  Eei 
into  components  along  and  normal  to  the  sliding,  getting  Ee2  as  the  normal  compon- 
ent. The  normal  component  of  the  velocity  of  Ex  must  be  the  same  as  Ee2  when 
paired  with  another  component  in  the  direction  of  the  sliding.  Hence  the  line 
drawn  to  the  intersection  of  e2ei  and  e4ea  represents  the  actual  velocity  of  Ex. 


FOUR-BAR  LINKAGE 


241 


The  instantaneous  axis  of  RHF  then  is  found  at  0.    The  velocity  of  R  is  therefore 
represented  by  Rn,  perpendicular  to  OR  and  of  such  length  that          =        ' 


FIG.  298 

Example  72.  Rate  of  Sliding  of  Gear  Teeth.  To  determine  the  rate  of  sliding  of 
one  gear  tooth  upon  another  at  any  position  it  will  be  necessary  to  find  the  linear 
velocity  of  the  point  of  contact  D  (Fig.  300)  in  each  of  the  teeth,  and  resolve  these 
linear  velocities  into  their  components  along  the  common  normal  and  the  common 
tangent.  Let  DE  represent  the  linear  velocity  of  D  around  Oij  the  components 
of  DE  along  the  common  normal  and  tangent  are  DF  and  DG  respectively.  The 


242 


ELEMENTS  OF  MECHANISM 

(fixed] 


FIG.  299 


FOUR-BAR  LINKAGE 


243 


direction  of  the  motion  of  D  around  02  is  along  the  line  DH.  To  find  the  magnitude 
of  its  linear  velocity  we  have  DF  as  its  component  along  the  common  normal,  since 
this  normal  is  the  line  of  connection  between  the  two  sliding  surfaces,  and  com- 


\ 


FIG.  300 

ponents  along  the  line  of  connection  must  be  equal.  This  will  give  DH  as  the  linear 
velocity  of  D  around  02,  and  DK  as  its  component  along  the  common  tangent.  The 
rate  of  sliding  will  be  found  to  be  GK,  equal  to  DG  +  DK,  since  the  components  along 
the  tangent  act  in  opposite  directions. 


CHAPTER  XI 
LINKWORK 

224.  A  Link  may  be  defined  as  a  rigid  piece  or  a  non-elastic  sub- 
stance which  serves  to  transmit  force  from  one  piece  to  another  or  to 
cause  or  control  motion. 

For  example,  that  part  of  a  belt  or  chain  running  from  the  driven  to 
the  driving  wheel,  the  connecting  rod  of  an  engine,  the  fluid  (if  assumed 
to  be  incompressible)  in  the  cylinder  of  a  hydraulic  press,  would  be 
links  according  to  the  above  definition.  In  ordinary  practice,  how- 
ever, the  name  is  applied  to  a  rigid  connector  (see  §  7),  which  may 
be  fixed  or  in  motion. 

225.  A  Linkage  consists  of  a  number  of  pairs  of  elements  con- 
nected by  links.     If  the  combination  is  such  that  relative  motion  of 
the  links  is  possible,  and  the  motion  of  each  piece  relative  to  the  others 
is  definite,  the  linkage  becomes  a  Kinematic  Chain.     If  one  or  more  of 
the  links  in  such  a  chain  be  fixed  the  chain  becomes  a  Mechanism 
(see  §  2). 

In  order  that  a  linkage  may  constitute  a  kinematic  chain,  the  num- 
ber of  fixed  points  or  points  whose  motions  are  determined  by  means 
outside  the  particular  linkage  in  question,  must  bear  such  a  relation 
to  the  number  of  links  in  the  linkage  that  the  linkage  may  form  a 
four-bar  linkage  or  a  combination  of  two  or  more  four-bar  linkages.  (See 
§  210.)  This  may  be  seen  by  reference  to  Figs.  301,  277  and  302. 


FIG.  301 

The  linkage  in  Fig.  301  consists  of  three  links  AB,  AC,  and  BC,  form- 
ing a  triangle  and  it  is  apparent  that  no  relative  motion  of  the  links 
can  occur  since  only  one  triangle  can  be  formed  from  three  given  lines. 

244 


LINKWORK 


245 


On  the  other  hand,  if  four  links  are  involved,  as  in  Fig.  277,  relative 
motion  of  a  definite  nature  will  result.  If  now  five  links,  as  AB,  BC, 
CD,  DE,  and  EA,  Fig.  302,  constitute  the  linkage,  any  link,  as  A E,  may 
be  fixed;  then  AB  and  ED  become  cranks,  but  a  given  angular  motion 
of  the  crank  A  B  does  not  impart  a  definite  resulting  angular  motion  to 


FIG.  302 

DE  unless  the  point  C  is  guided  by  some  external  means.  If,  how- 
ever, C  is  guided  by  the  crank  FC  turning  about  the  fixed  center  F  the 
motions  of  all  the  links  become  determinate.  But  the  linkage,  by  the 
addition  of  the  crank  FC,  has  now  been  transformed  into  a  combination 
of  two  four-bar  linkages,  namely:  ABCF  and  FCDE  with  A,  F,  and  E 
fixed. 

In  general,  it  may  be  said  that  any  mechanism  may  be  analyzed  as  a 
four-bar  linkage  or  as  a  combination  of  two  or  more  such  linkages. 

226.  The  Four-Bar  Linkage.  In  §  210  a  four-bar  linkage  was  de- 
scribed and  special  names  were  given  to  the  links  when  one  of  them  was 
fixed.  It  is  not  necessary  that  any  link  be  absolutely  at  rest.  The 
link  AD,  Fig.  277,  which  is  there  assumed  to  be  fixed,  may  be  attached 
to  some  other  part  of  the  machine  which  itself  is  in  motion.  ABCD 
remains  a  four-bar  linkage  and  the  relative  motions  of  its  four  links  are 
unchanged  although,  of  course,  the  absolute  motion  of  each  link  de- 
pends not  only  upon  its  motion  with  relation  to  the  link  (in  this  case 
AD)  which  is  assumed  to  be  fixed,  but  also  upon  the  motion  which 
that  link  has. 

In  view  of  that  which  has  preceded,  a  four-bar  linkage  may  be  de- 
scribed as  consisting  of  four  cylindrical  pairs  of  elements,  or  their  equiv- 
alent, each  element  of  each  pair  being  connected  by  a  rigid  link,  or  its 
equivalent,  to  one  of  the  elements  of  an  adjacent  pair.  Thus,  the  pin 
A,  Fig.  272,  is  connected  by  the  fixed  frame  E  to  the  pin  D.  The  en- 
larged end  of  the  crank  F  which  contains  the  cylindrical  hole  paired 
with  A  is  connected  to  one  of  the  pair  at  C,  and  so  on. 


246  ELEMENTS  OF  MECHANISM 

227.  Relative  Motion  of  the  Links  in  a  Four-Bar  Linkage.     Since, 
as  shown  in  §  226,  the  motions  of  the  links,  relative  to  some  one  link 
assumed  to  be  fixed,  are  not  changed  if  motion  is  imparted  to  that  link, 
it  follows  that  the  motion  of  any  link,  relative  to  any  other  link  of  the 
linkage,  is  the  same  whichever  link  is  fixed.     In  other  words,  the  rela- 
tive motions  of  the  links  of  a  four-bar  linkage  are  independent  of  the  fixed- 
ness of  the  links.    This  principle  is  taken  advantage  of  in  the  applica- 
tion of  four-bar  linkages,  particularly  in  cases  where  centrodes  are 
substituted  for  some  of  the  links,  as  will  be  illustrated  later. 

The  laws  relating  to  the  motions  of  the  links  can  be  studied  more 
conveniently  by  assuming  one  link  fixed,  and  this  method  will  be 
followed  in  the  succeeding  paragraphs. 

228.  Angular  Speed  Ratio  of  Cranks.     The  angular  speeds  of  the 
two  cranks  of  a  four-bar  linkage  are  inversely  as  the  lengths  of  the  perpen- 
diculars or  any  two  parallel  lines  drawn  from  the  fixed  centers  to  the  center 
line  of  the  connecting  rod;  also,  inversely  as  the  distances  from  the  fixed 
centers  to  the  point  of  intersection  of  the  center  line  of  the  connecting  rod 
and  the  line  of  centers  (produced  if  necessary). 

This  law  may  be  shown  to  be  true  in  two  ways. 
1°.  By  reference  to  the  instantaneous  axis  of  the  connecting  rod. 
Let  ABCD  (Fig.  303)  represent  the  linkage,  AD  being  the  fixed  link. 
To  show  that 

Angular  speed  of  DC  _  Am  _Ak 

Angular  speed  of  AB      Dn      Dk 


?T 


The  instantaneous  axis  of  BC  is  at  0  and  since  the  linear  speeds  of 
B  and  C  are  to  each  other  as  their  distances  from  the  instantaneous 

axis  'see  §  219) 

Linear  speed  C  _  OC  ,.,, 

Linear  speed  B  =  OB* 


LINKWORK  247 

But  since  the  angular  speed  of  DC  is  equal  to  the  linear  speed  of  C  di- 
vided by  the  radius  DC  (see  §  38^  it  follows  that 

,  -^      Linear  speed  C  /TTX 

Angular  speed  DC  =  sd? (II) 

.L/C/ 

Similarly 

,  A  n      Linear  speed  B  /TTTX 

Angular  speed  AB  = ,£ (III) 

Dividing  (II)  by  (III) 

Linear  speed  C 

Angular  speed  DC DC Linear  speed  C      AB     ,-._,, 

Angular  speed  AB      Linear  speed  B      Linear  speed  B      DC 

AB 

Whence,  combining  (I)  and  (IV), 

Angular  speed  DC      PC      AB  _  PC       AB 
Angular  speed  AB      OB  X  DC      DC  X  OB 

Draw  Ot,  Dn  and  Am  perpendicular  to  BC.    Then,  from  the  similar 
triangles  OtC  and  DnC} 

Dn  =  DC' 

a.    .,    ,  Am      AB 

Similarly  W  =  QB 

Substituting  these  values  in  (V) 

Angular  speed  DC       Qt_      Am  _  Am  ,„  . 

Angular  speed  AB  ~  Dn      ~0t  ~~  ~Dn" 

Producing  CB  to  meet  DA  at  k,  it  will  be  noticed  that  triangles 
Dnk  and  Amk  are  similar,  hence 

Ak  _  Am 
Dk  =  Dn" 


Therefore,  from  equation  (73), 

Angular  speed  DC  _  Ak 
Angular  speed  A  B      Dk 


(74) 


In  determining  the  speed  ratio  of  the  cranks  for  any  specific  case 
either  equation  (73)  or  (74)  may  be  used  as  happens  to  be  more  con- 
venient. 


248  ELEMENTS  OF   MECHANISM 

2°.  By  resolution  of  velocities.  In  Fig.  304  let  Bbi  represent  the 
linear  velocity  of  B.  Then  Bbz  is  the  component  of  this  velocity  along 
CB  and  Cci  is  the  velocity  of  C. 


FIG.  304 

rpr      f  Linear  speed  C      Cci 

Therefore  ^  -  v     ,  p  =  -^  » 

Linear  speed  B      Bbi 

and  since  linear  speed  is  equal  to  angular  speed  divided  by  the  radius, 
and  the  triangles  CciCz  and  DCn  are  similar  as  are  also  Bbibz  and  ABm, 
this  may  be  written 

Angular  speed  DC  __  Cci  _._  Bbi  _  Cc%      Am 
Angular  speed  AB  ~  ~DC  '  ~  Z#  ~  Dn      Bb~2' 


But  Cc2 

Hence  the  equation  becomes 

Angular  speed  DC  _  Am 
Angular  speed  AB  ~  Dn  ' 
as  in  (73). 

This  angular  speed  ratio  of  course  varies  for  every  relative  position 
of  the  links;  but  if  the  perpendicular  from  the  instantaneous  axis  to 
the  center  line  of  the  connecting  rod  should  fall  at  the  intersection  of 
the  center  line  of  the  connecting  rod  and  the  line  of  centers,  that  is,  in 
Fig.  303,  if  the  points  t  and  k  should  coincide,  the  angular  speed  ratio 
is  essentially  constant  for  slight  movements  in  either  direction.  The 
same  would  be  true  should  the  points  B  and  C  be  moving  in  lines  par- 
allel to  each  other. 

229.  Diagrams  for  Representing  Changes  in  the  Linear  Speed 
Ratio  or  Angular  Speed  Ratio  in  Any  Linkage.  To  obtain  a  clear 
knowledge  of  the  change  in  velocity  ratio  in  any  linkage  a  diagram  may 
be  drawn  where  the  abscissae  may  represent  successive  positions  of  one 
of  the  oscillating  links,  and  the  ordinates  represent  the  angular  speed 
ratio  of  the  oscillating  links.  A  smooth  curve  through  the  points  thus 
found  would  show  clearly  the  fluctuations  in  the  angular  speed  of  one  of 
the  links  relative  to  the  other.  A  curve  for  linear  speed  ratio  could  be 
similarly  plotted. 


LINKWORK 


249 


In  the  linkage  shown  in  Fig.  305  let  A B  turn  uniformly  left  handed ;  re- 

, .     angular  speed  DC  f 

quired  a  curve  to  represent  the  ratio  -    — -, — ,   .  n  for  a  complete 

angular  speed  A  B 

rotation  of  AB.    Take  positions  of  AB  at  intervals  of  30°  and  draw  per- 


FIG.  305 

pendiculars  from  D  and  A  to  BC  in  each  of  its  positions.  The  ratio  of  the 
two  perpendiculars  in  each  position  will  give  the  angular  speed  ratio : 
thus,  starting  with  AB  as  o.4 
given  in  the  figure, 

angular  speed  DC  _  ~ 

angular  speed  AB  ~ 
in  the  position  ABiCiD 
we  have 

angular  speed  DC  _  Ami . 
angular  speed  AB     Dni ' 
etc.  Plotting  these  values 
as  ordinates  and  the  30° 
positions  of  AB  as  ab-   0.3 
scissse  will  give  the  curve 
shown  in  Fig.  306.     Ordi- 
nates above  the  zero  line 


U.4 

0.3 
0.2 
O.I 
0 
01 
0.2 
0.3 
0.4 
05 

/ 

' 

\ 

' 

1 

\ 

\ 

1 

\ 

3 

Ou  6 

)u  y 

<J»  li 

0"li 

0"  ll 

V 

0»2< 

0»2i 

0»3C 

0»3i 

i 

x 

\ 

I 

\ 

i 

Ord 

mat( 

,c_  Angul 

ar  S/ 

eed  L, 

.c, 

\ 

/ 

Angular  <S/ 

eed  A 

.B, 

Abs 

c!ss' 

.- 

Positiom 

Of 

A,B. 

^ 

f 

indicate  that  the  cranks 

turn  in  the  same  direction,  while  those  below  show  that  the  directions 

of  rotation  are  opposite. 

230.   Dead  Points.     A  position  in  the  cycle  of  motion  of  the  driven 
crank  of  a  linkage  in  which  it  is  in  line  with  the  connecting  rod,  and 


250 


ELEMENTS  OF   MECHANISM 


therefore  cannot  be  moved  by  the  connecting  rod  alone,  is  known  as 
a  dead  point.  If  the  driven  crank  makes  complete  revolutions  there 
are  two  such  positions  in  its  cycle. 

231.  Crank  and  Rocker.  Let  the  link  AD  (Fig.  307)  be  fixed,  and 
suppose  the  crank  AB  to  revolve  while  the  lever  DC  oscillates  about 
its  axis  D.  In  order  that  this  may  occur,  the  following  conditions  must 
exist. 


1° 

2° 


AB  +  BC  +  DO  AD, 
AB  +'AD]+  DC  >  AB, 


3°  AB  +  BC  -  DC  <  AD, 
4°  BC  -AB  +  DOAD. 


1°  and  2°  must  hold  in  order  that  any  motion  shall  be  possible; 
3°  can  be  seen  from  the  triangle  AC%D  in  the  extreme  right  position 


FIG.  307 

,  which  must  not  become  a  straight  line;   and  4°  can  be  seen 
from  the  triangle  ACiD,  in  the  left  extreme  position  ABiCJ). 

There  are  two  points  Ci  and  €2  in  the  path  of  C  at  which  the  motion  of 
the  lever  is  reversed,  and  it  will  be  noticed  that  if  the  lever  DC  is  the 
driver,  it  cannot,  unaided,  drive  the  crank  AB,  as  a  pull  or  a  thrust  on 
the  rod  BC  would  only. cause  pressure  on  A,  when  C  is  at  either  d  or 
£2.  If  AB  is  the  driver,  this  is  not  the  case. 

The  above  form  of  linkage  is  applied  in  the  beam  engine  as  shown 
in  Fig.  308,  the  fixed  link  AD  being  formed  by  the  engine  frame;  corre- 
sponding parts  are  lettered  the  same  as  in  Fig.  307.  The  instantaneous 
axis  is,  for  the  position  shown,  at  0,  and  for  the  instant  the  linear  speed 
of  B  is  to  the  linear  speed  of  C  as  OB  is  to  OC,  or  as  Bf  is  to  Ce,  the 
line  efj  drawn  parallel  to  BC,  being  made  use  of  when  the  point  0  comes 
beyond  the  limits  of  the  drawing. 

The  angle  through  which  the  lever  DC  (Fig.  307)  swings  can  be 
calculated  for  known  values  of  AB,  BC,  CD,  and  DA. 

From  the  triangle  ACzD 

DC2  +  AD*  -  (BC  +  AB)2 


cos  ADC2  = 


2  (DC  X  AD) 


LINKWORK 


251 


DC2  +  AD2  -  (BC  -  AB)* 


and  from  the  triangle  ACiD 

cos  ADC,  =  2  (DC  X  AD) 

and  CiDC2  =  ADC2  -  ADCi.  (75) 

Thus  the  two  angles  ADCz  and  ADCi  can  be  calculated,  and  their 
difference  will  give  the  angle  required. 


FIG.  308 


If  tne  link  BC  is  made  stationary,  the  mechanism  is  similar,  the  only 
difference  being  in  the  relative  lengths  of  the  connecting  rod  and 
stationary  piece  or  frame. 


FIG.  309 

Fig.  309  shows  a  case  of  the  crank  and  rocker  as  applied  in  wool- 
combing  machinery.  Here  the  crank  A B  turns  uniformly  on  its  axis  A, 
while  CD  oscillates  about  D;  both  axes  A  and  D  are  attached  to  the 


252 


ELEMENTS  OF  MECHANISM 


frame  of  the  machine,  which  forms  the  fixed  link  AD.  The  connecting 
rod  CB  is  prolonged  beyond  B,  and  carries  a  comb  E  at  its  extremity, 
which  takes  a  tuft  of  wool  from  the  comb  F  and  transfers  it  to  the  comb 
G,  both  combs  F  and  G  being  attached  to  the  frame  of  the  machine. 
The  full  lines  show  the  position  of  the  links  when  the  comb  E  is  in  the 
act  of  rising  through  the  wool  on  F,  thus  detaching  it,  and  the  dotted 
lines  show  the  position  of  the  links  when  the  comb  EI  is  about  to  deposit 
the  tuft  of  wool  on  G.  The  same  combination  inverted  is  used  in  some 
forms  of  wool-washing  machines. 

232.   Drag  Link.     If  the   link   A  B,   in  the  linkage  Fig.    307,   is 
made  the  stationary  piece  or  frame,  as  in  Fig.  310,  the  links  BC  and 


FIG.  310 

AD  revolve  about  B  and  A  respectively,  that  is,  become  cranks,  and 
CD  becomes  a  connecting  rod.  This  mechanism  is  known  as  the  drag- 
link. 

In  order  that  the  cranks  may  make  complete  revolutions,  and  that 
there  may  be  no  dead-points,  the  following  conditions  must  hold: 

1°  Each  crank  must  be  longer  than  the  line  of  centers,  which  needs 
no  explanation. 

2°  The  link  CD  must  be  greater  than  the  lesser  segment  C4/  and  less 
than  the  greater  segment  CJ)^  into  which  the  diameter  of  the  greater 
of  the  two  crank  circles  is  divided  by  the  smaller  circle.  This  may  be 
expressed  as  follows: 


CD  >  AB  +  AD  -  BC  (see  triangle 

CD  <  AD  +BC  -  AB  (see  triangle  BCzDz). 


LINKWORK 


253 


Producing  the  center  line  of  the  connecting  rod  until  it  intersects  the 
line  of  centers  at  e,  and  dropping  the  perpendiculars  Am  and  Bn  upon  it, 

Angular  speed  AD  _  Be  _  Bn 
Angular  speed  BC      Ae      Am 

In  the  positions  ABCiDi  and  ABCsDs,  when  CD  is  parallel  to  the 
line  of  centers,  the  angular  speeds  of  AD  and  BC  are  equal,  since  the 
perpendiculars  Bn  and  Am  then  become  equal. 

If  BC  revolves  left-handed  and  is  considered  the  driver,  it  will  be 
noticed  that  between  the  positions  ABCSDS  and  ABCiDi  the  crank  AD 
is  gaining  on  BC,  and  between  ABCiDi  and  ABC&Ds  it'is  falling  behind 
BC. 

Fig.  311  shows  an  application  of  the  drag  link  for  driving  the  ram  of 
a  Dill  slotter.  The  links  in  this  figure  are  lettered  to  correspond  with 
Fig.  310.  The  large  gear,  turning  on  a  fixed  boss,  centered  at  B  on  the 


FIG.  311 

frame,  carries  the  pin  C  and  forms  the  driving  crank  corresponding  to 
BC,  Fig.  310.  The  shaft  A  has  its  bearing  in  a  hole  in  the  large  boss 
on  which  the  gear  turns  and  has  keyed  to  it  the  crank  arm  AD.  On 
the  other  end  of  this  shaft  is  another  crank  arm,  or  its  equivalent,  the 
center  line  of  which  is  AN.  To  this  latter  crank  arm  is  attached 
the  connecting  rod  which  drives  the  ram.  The  mechanism  is  shown 


254 


ELEMENTS  OF  MECHANISM 


in  the  position  which  it  occupies  when  the  ram  is  about  at  the  middle 

of  the  downward  or  cutting  stroke. 

233.   The  Double  Rocking  Lever  (Fig.  312)  shows  the  same  linkage 

with  DC  as  the  fixed  link.     In  this  case  the  cranks  CB  and  DA  merely 

oscillate  on  their  axes  C  and 
D.  The  extreme  positions 
may  be  assumed  at  AiBi  and 


234.  Parallel  Crank  Four- 
Bar  Linkage.  In  Fig.  313, 
the  crank  AB  is  equal  in 
length  to  the  crank  CD  and 
the  line  of  centers  AD  is 
equal  to  the  connecting  rod 
BC.  The  center  lines  of  the 
linkage  thus  form  a  parallelo- 
gram in  every  position,  provided  the  cranks  turn  in  the  same  direction. 
Therefore,  the  perpendiculars  Am  and  Dn  are  always  equal  and  the 
two  cranks  are  always  turning  at  the  same  angular  speed.  A  familiar 
example  of  this  linkage  is  furnished  by 
the  cranks  and  parallel  rod  of  a  loco- 
motive. In  this  case  the  link  formed 
by  the  center  line  of  bearings  in  the 
frame  carrying  the  axles  of  the  two 
driving  wheels  corresponds  to  the  line 
of  centers,  but  is  itself  in  motion. 

235.  Non-Parallel  Equal  Crank  Linkage. 


Fixed  Axis 


Fixed  Axis 
D 


Fixed  Axis 

FIG.  313 

In  the  linkage  shown  in 
Fig.  314,  AB  is  equal  to  CD  and  AD  is  equal  to  BC.  Provision  is  made, 
however,  to  cause  the  cranks  to  turn  in  opposite  directions;  in  which 
case  the  perpendiculars  Am  and  Dn  do  not  remain  equal  to  each  other. 

Therefore,  if  the  crank  AB  turns  with 
uniform  angular  speed,  the  crank  DC  has 
a  varying  angular  speed,  although  both 
make  one  complete  turn  in  the  same 
length  of  time. 

The  opposite  directions  of  revolution 
may  be  secured  by  providing  some 
means  of  causing  the  cranks  to  pass  the  dead  points  in  the  proper 
direction.  This  may  be  accomplished  by  means  of  some  device  placed 
at  the  instantaneous  axis  of  the  connecting  rod  when  in  these  positions 
although  the  entire  linkage  may  be  replaced  by  gears  whose  pitch  sur- 
faces are  the  centrodes  of  two  of  the  opposite  links.  If  in  the  linkage 


•      A. 
Fixed  Axis 


FIG.  314 


LINKWORK 


255 


in  Fig.  314,  AD  being  the  line  of  centers,  the  centrode  of  BC  is  drawn 
it  will  be  found  to  be  the  hyperbola,  part  of  which  is  shown  in  full  lines 
in  Fig.  315,  with  foci  at  A  and  D  and  transverse  axis  fg  lying  along  the 


FIG.  315 

fixed  link  AD  and  equal  in  length  to  AB  =  DC.  Similarly,  if  BC  is 
considered  as  the  line  of  centers  the  centrode  of  AD  is  the  hyperbola, 
part  of  which  is  shown  in  dotted  lines  in  Fig.  315,  with  foci  at  B  and  C 


FIG.  316 


and  transverse  axis  hk  lying  along  BC  and  equal  in  length  to  AB  =  DC. 
If,  with  A  and  D  as  the  fixed  centers,  a  pin  is  placed  on  the  connecting 
rod  BC  at  h  (Fig.  316)  and  a  corresponding  open  eye  at  g  on  the  fixed 


256  ELEMENTS  OF  MECHANISM 

line  AD,  the  points  h  and  g  being  the  same  points  as  h  and  g  in  Fig.  315, 
the  pin  at  h  will  mesh  into  the  eye  at  g  when  the  linkage  reaches  one 
dead  point  and  thus  insure  that  the  cranks  pass  the  dead  point  in  the 
proper  direction.  A  similar  pin  at  k  and  eye  at  /  will  provide  for 
passing  the  other  dead  point. 

If  the  link  DC  is  assumed  to  be  the  fixed  link  the  centrode  of  AB  is 
the  ellipse  N,  Fig.  317,  with  foci  at  D  and  C  and  the  length  of  the  major 


FIG.  317 

axis  is  equal  to  AD=  BC.  Similarly  if  AB  is  assumed  to  be  fixed  the 
centrode  of  DC  is  the  ellipse  M  with  A  and  B  as  foci  and  with  AD  =  BC 
as  the  length  of  the  major  axis. 

The  ellipses  being  closed  curves,  it  is  possible  to  substitute  for  the 
links,  short  cylinders  whose  bases  are  of  the  form  of  the  elliptical  cen- 
trodes.  Let  it  be  supposed  that  two  equal  gears  are  made  whose 
pitch  surfaces  are  of  the  form  of  the  ellipses  shown  in  Fig.  317, 
and  that  these  pitch  ellipses  are  placed  in  contact  with  each  other  as 
in  the  figure.  Links  BC  and  DA  may  be  attached  to  the  gears  at 
the  foci  of  the  ellipses,  and  if  the  gear  N  is  held  still  and  M  rolled 
around  it  the  links  CB  and  DA  will  turnabout  the  centers  C  and 
D  respectively  and  the  line  joining  the  foci  A  and  B  will  have  the  same 
motion  that  it  would  have  in  the  linkage.  Or  M  may  be  pivoted  at  the 
focus  A  and  N  at  the  focus  D  and  both  allowed  to  turn,  the  result  being 
the  same  as  in  the  linkage  with  AD  as  the  fixed  link,  AB  and  DC  as 
cranks. 

If  the  cranks  are  prolonged  as  shown  in  Fig.  318  and  pins  placed  at 
n  and  m\  with  corresponding  eyes  at  m  and  n\,  these  points  being  the 
same  as  the  points  with  corresponding  letters  in  Fig.  317,  a  means 
is  provided  for  passing  the  dead  points  if  the  links  themselves  are 
used. 

If  desired,  one  dead  point  may  be  passed  by  means  of  pin  and  eye  on 
the  elliptical  centrodes  and  the  other  dead  point  by  pin  and  eye  on  the 
hyperbolic  centrodes.  Or  a  few  teeth  on  sectors  of  gears  at  these  parts 
of  the  centrodes  might  be  used  in  place  of  the  pin  and  eye. 


LINKWORK 


257 


Elliptic  gears  designed  on  the  above  principles  have  been  used  to 
drive  the  rams  of  machine  tools,  such  as  slotters,  so  as  to  give  a  slow 
cutting  stroke  to  the  tool,  and  a  quicker  return  stroke.  In  applying 


FIG.  318 

the  gears  for  such  a  purpose  one  of  them,  as,  for  instance,  M",  is  on  a 
shaft  at  A  driven  at  a  uniform  speed  from  some  external  source  of 
power.  The  other  gear  AT  is  on  a  shaft  at  D  to  which  is  attached  the 
crank  or  other  device  for  moving  the  ram. 

236.  Slow  Motion  by  Linkwork.  The  four-bar  linkage  can,  if 
properly  proportioned,  be  made  to  produce  a  slow  motion  of  one  of  the 
cranks.  Such  a  combination  is  shown  in  Fig.  319,  where  two  cranks 


FIG.  319 


AB  and  DC  are  arranged  to  turn  on  fixed  centers  and  are  connected 
by  the  link  BC.  If  the  crank  AB  is  turned  right-handed,  the  crank 
DC  will  also  turn  right-handed,  but  with  decreasing  speed,  which  will 
become  zero  when  the  crank  A B  reaches  position  ABi  in  line  with  the 
link  BCi  any  further  motion  of  AB  will  cause  the  link  DC  to  return 


258 


ELEMENTS  OF  MECHANISM 


toward  its  first  position,  its  motion  being  slow  at  first  and  then  gradually 
increasing.  This  type  of  motion  is  used  in  the  Corliss  valve  gear,  as 
shown  in  Fig.  320.  The  linkage  ABCD,  moving  one  of  the  exhaust 


FIG.  320 

valves,  will  give  to  the  crank  DC  a  very  slow  motion,  when  C  is  near  Ci, 
when  the  valve  is  closed,  while  between  C  and  €2,  when  the  valve  is 
opening  or  closing,  the  motion  is  much  faster.  The  same  is  true  for 
the  admission  valves,  as  shown  by  the  linkage  AEFG. 

237.  Linkwork  with  a  Sliding  Pair.  In  Chapter  X  (§211)  there 
was  shown  the  relation  between  a  linkage,  such  as  the  one  in  Fig.  321, 
and  the  simple  four-bar  linkage.  It  is  important  that  this  relation  be 
clearly  understood  before  proceeding  and  it  is  suggested  that  the  reader- 
review  that  discussion  before  studying  what  follows. 


At  laSofty 

FIG.  321 

Referring  now  to  Fig.  321,  the  four  links  of  the  linkage  are  AB,.BC, 
CD  and  AD,  the  lines  AD  and  CD  meeting  at  infinity,  that  is,  being 
parallel,  and  perpendicular  to  the  center  line  of  the  slot  in  T.  It  should 
be  borne  in.  mind  that  the  piece  T  is  not  one  of  the  links  of  the  four-bar 


LINKWORK 


259 


linkage,  but  that  T  and  the  block  pivoted  to  the  link  BC,  replace,  and 
constitute  the  equivalent  of,  the  two  infinite  links  AD  and  CD. 

Four  distinct  mechanisms  may  be  obtained  from  this  form  of  the 
linkage  by  making  each  one  of  the  links,  in  turn,  the  stationary  link. 
These  four  mechanisms,  with  applications,  will  now  be  discussed. 

238.  The  Sliding-Block  Linkage.  Considering  the  piece  T  (Fig. 
321)  as  fixed  gives  the  mechanism  shown  in  Fig.  322  which  is  the 


(Fixed  Axis) 


FIG.  322 

mechanism  commonly  used  in  pumps  and  direct-acting  steam  engines. 
When  employed  in  a  steam  engine,  the  block  at  C,  called  the  cross- 
head,  is  the  driver  and  the  crank  AB  the  follower;  in  a  pump  the 
reverse  is  the  case. 


FIG.  323 

Movement  of  Crosshead.     In  Fig.  323  let  AB  represent  the  crank,  BC 
the  connecting  rod,  and  mn  the  path  of  the  point  C  in  the  crosshead. 


260 


ELEMENTS  OF  MECHANISM 


The  travel  of  the  crosshead  mn  is  equal  to  twice  the  length  of  the 
crank  AB  and  the  distance  of  C  from  A  varies  between  BC  -\-  AB 
=  An  and  BC  —  AB  =  Am,  AB  being  the  length  of  the  crank,  and 
BC  the  length  of  the  connecting  rod. 

To  find  the  distance  the  point  C  has  moved  from  n,  the  beginning 
of  its  stroke  or  travel,  let  the  angle  made  by  the  crank  with  the  line 
An  be  represented  by  0,  and  draw  Bg  perpendicular  to  An.  The  move- 
ment of  the  crosshead  from  the  beginning  of  its  stroke  is,  for  the  angu- 
lar motion  6  of  the  crank, 

Cn  =  An  —  AC  —  An  —  (Ag  +  gc). 
From  the  right  triangle  BCg 


Hence 


-  Bgz. 


Cn  =  An  —  AB  cos  0  — 

=  AB  +  BC  -  AB  cos  0  -  VSC2  -  AB2  sin2  0, 


=  AB  (1  -  cos  0)  +  BC 


1  - 


sin20 


(76) 
(77) 


It  will  be  noticed  that  Equation  (77)  indicates  that  the  displacement 
differs  from  that  which  C  would  have  if  its  motion  were  harmonic 
(assuming  AB  to  turn  with  uniform  speed)  by  the  term 


£C     1  - 


-^sin20 
BC2 

(See  Equation  5,  page  7.) 

and  that  the  value  of  this  term  decreases  as  BC  increases  relative  to  AB. 

That  is,  the  longer  the  connecting  rod  is  made  relative  to  the  crank  the 

more  nearly  the  motion  of  the  cross- 
head  approaches  harmonic  motion. 
The  motion  may  be  represented 
graphically    by    plotting   a    curve, 
where  the  ordinates  represent  suc- 
cessive values  of  Cn,  and  the  abscissae 
represent  angular  positions  of  the 
crank  AB.     Fig.  324  shows  the  curve 
for  the  linkage  given  in  Fig.  328,  the 
120°    150°    isoa    dotted  line  being  the  corresponding 
curve,  if  the  motion  of  C  were  har- 
monic. 
If  a  device  similar  to  that  shown  in  Fig.  325  is  used,  in  which  a  rigid 

rod  is  attached  to  the  crosshead,  this  rod  having  a  slot  in  it  at  right 


30° 


60* 


FIG.  324 


LINKWORK 


261 


angles  to  the  line  of  motion  of  the  crosshead  and  embracing  a  block 
pivoted  on  the  crank  pin,  then  the  crosshead  would  have  harmonic 
motion  if  the  crank  turned  at  uniform  speed. 


/B 

Connecting  Rod  of, 

"  fc  In  fin  it 


FIG.  325 

This  device  is  the  equivalent  of  a  connecting  rod  of  infinite  length, 
and  the  whole  mechanism  may  be  thought  of  as  a  four-bar  linkage  in 
which  only  one  of  the  links,  namely,  the  crank  AB,  is  a  finite  quantity. 
The  line  of  centers  and  the  infinite  connecting  rod  are  indicated  in  the 
figure.  The  other  crank  is  an  imaginary  line  parallel  to  the  line  of 
centers  at  an  infinite  distance  away. 


Crosshead  Pin 
C 


FIG.  326 


Linear  Speed  of  Crosshead.  It  is  convenient  to  be  able  to  determine 
the  velocity  of  the  crosshead,  and  hence  of  the  piston,  of  a  steam  engine 
for  different  positions  of  the  stroke  when  the  speed  of  the  crank  pin  is 
known.  In  Fig.  326 

Linear  speed  of  C  _  OC 

Linear  speed  of  B      OB 


(78) 


262  ELEMENTS  OF  MECHANISM 

Through  A  draw  a  line  perpendicular  to  the  center  line  of  the  slot,  and 
extend  the  center  line  of  the  connecting  rod  to  cut  this  line  at  m.  Then 
the  triangles  OCB  and  mBA  are  similar.  Hence, 

OC  =  Am 
OB      AB  '- 

Substituting  this  in  Eq.  (78)  gives 

Linear  speed  of  C  _  Am  .„  . 

Linear  speed  of  B  ~  AJS' 

or  in  words,  The  linear  speed  of  the  crosshead  or  piston  of  a  steam  engine 
is  to  the  linear  speed  of  the  crank  pin  as  the  distance  between  the  crank 
shaft  and  the  point  where  the  connecting  rod  cuts  the  perpendicular  through 
the  center  of  the  crank  shaft  is  to  the  length  of  the  crank. 
From  the  similar  triangles  CAm  and  CgB 

Am      AC  AC  A    +   C 


Whence 

AB  sin  6  {AB  cos  6  +  V#C2  -ZB2sin20} 
Am  =  --  [  -  ,  -  -• 

VBC2  -  Atf  sin2  6 
Substituting  this  value  in  (79)  gives 

Linear  speed  of  C        .    _   .        A  B  sin  0  cos  0  /0/vv 

—£-&  =  sin  6  +     >  •  (80) 

Linear  speed  of  B  A/£C2  _  Z82  sin2  0 

This  same  result  may  be  derived  by  another  method.  Let  v  repre- 
sent the  speed  of  the  crosshead,  s  its  displacement,  and  t  the  time  dur- 
ing which  the  displacement  has  taken  place. 

Then  v  =  j|-    (See  Equation  (2),  §  25.) 

Letting  o>  represent  the  angular  speed  of  A  B  in  radians  per  unit  of 
time  and  expressing  6  as  at,  Equation  (76)  may  be  written 


s  =  AB  +  BC  -  AB  cos  ut  -  V  £C2  -  AB2  sin2  <*t. 
Therefore 


j    en  A  0  .  sin  ut  cos  ut        /01, 

Linear  speed  of  C  =  -T.  =  wAB  sin  ut  -\  --  ,  -  •    (81) 

dt  V£C2-ZB2sin2o>* 

But  the  linear  speed  of  B  =  uAB. 

Linear  speed  of  C:       .  A  B  sin  at  cos  wt  /0  , 

Therefore  ^—  ,    £  p  =  sin  ut  H  --  ,  •       (82) 

Linear  speed  of  B  V^C2  -  AB2  sin2  ut 


LINKWORK 


263 


When  6  =  90°,  Am  =  AB  and  the  speeds  of  C  and  B  are  equal.  To 
find  other  values  of  6,  when  C  and  B  have  equal  speeds;  from  Equa- 
tion (80), 

.    .   .        AB  sin  6  cos  9 
1  =  sm  6  + 


V  BC2  -  AB2  sin2  0 
Solving  this  for  sin  6  gives 

Sin  0  =  -£^  {  -BC  ±VsI52 


BC2} . 


(83) 


The  linear  speed  ratio  between  C  and  B  may  be  shown  graphically, 
using  coordinate  axes,  the  ordinates  representing  the  ratio  and  the  ab- 
scissae representing  angular  positions  of  the  crank  AB.  Fig.  327  shows 
the  curve  for  the  linkage  given  in  Fig.  328. 


°0°        30°       60°  75° 90°       120°      '50°      180" 

FIG.  327 

Fig.  328  illustrates  other  methods  of  showing  the  linear  speed  ratio. 
In  this  figure  the  constant  linear  speed  of  B  is  represented  by  the  crank 
length  AB.  From  Equation  (79) 

Linear  speed  C  _  Am 

Linear  speed  B      AB 

Therefore,  by  laying  off  on  the  line  AB,  which  shows  the  crank  posi- 
tion, the  distance  At  =  Am,  and  repeating  this  construction  for  a 


FIG.  328 

sufficient  number  of  crank  positions,  the  full  curve  AtA  will  be  obtained 
where  the  intercept  At  on  the  crank  line  shows  the  velocity  of  C,  AB 
being  the  constant  velocity  of  B.  A.  similar  curve  would  be  found  for 


264 


ELEMENTS  OF  MECHANISM 


the  crank  positions  below  the  line  MA.  Similarly  the  full  curve  Nt\M 
might  be  obtained  by  laying  off  on  the  successive  perpendiculars  drawn 
through  the  point  C  the  corresponding  distances  Am.  The  dotted  curves 
are  the  corresponding  curves  for  harmonic  motion  and  are  circular. 

If,  in  Fig.  326,  the  crank  AB  is  the  driver,  it  can  always  produce 
reciprocating  motion  in  the  block  at  C;  but  if  the  block  is  the  driver,  it 
cannot  produce  continuous  circular  motion  unless  some  means  of  pass- 
ing the  dead  points  be  employed.  This  is  usually  accomplished  in 
steam  engines  by  attaching  to  the  crank  shaft  a  heavy  fly  wheel,  the 
momentum  of  which  carries  the  crank  by  the  dead  points.  The  im- 
possibility of  starting  at  the  dead  points  still  remains. 

To  obviate  this  difficulty  two  crank  and  connecting-rod  mechanisms 
may  be  combined,  as  shown  in  Fig.  329,  where  the  cranks  are  placed 


FIG.  329 

at  right  angles  to  each  other  and  joined  by  a  shaft.  This  combination 
is  employed  in  locomotives,  and  in  hoisting  and  marine  engines,  one 
crank  being  very  near  its  best  position  to  be  acted  on  by  the  rod  while 
the  other  is  at  a  dead  point. 

Fig.  330  shows  another  method  of  passing  the  dead  points  sometimes 
used  in  marine  engines.  Here  the  two  connecting  rods  BC  and  BCi 
are  located  in  parallel  planes  and  act  upon  the  same  crank  AB.  By 
suitably  forming  the  ends  of  the  rods,  they  might  be  located  in  the 
same  plane. 

Acceleration  of  Crosshead.    Since  (from  Equation  (4))  a  =  ^>  where 

a  represents  the  acceleration,  v  the  linear  speed  and  t  the  time,  Equation 
(81)  may  be  differentiated  giving 


d  (    .  ~  coAB2  sin  coZ  cos  co£   \ 

-37  I  uAB  sin  cot  -\  --  ,  I 

*\  V^2-ZB2sin2arf/ 


=  u?AB  cos  ut  H — 7= 


LINKWORK 


cos2  ut  —  sin2  at)  .    w2AB*  sin2  col  cos2  ut 
(BC2  -  AB2  sin2  utf       (BC2  -  AB2  sin2  co^)^ 


265 


(84) 


The  acceleration  may  be  found  with  a  fair  degree  of  accuracy  graphi- 
cally by  plotting  a  velocity  curve  as  shown  in  Fig.  331  in  which  the 
abscissae  are  linear  distances  representing  time  (for  example,  each  space 
may  represent  one  second)  and  the  ordinates  are  the  corresponding 


FIG.  330 

velocities  of  C.  Then  if  a  tangent  be  drawn  at  any  point  as  k  and  this 
tangent  is  made  the  hypotenuse  of  a  right  triangle  kpd  whose  base  is 
parallel  to  xx  and  equal  to  the  distance  which  represents  one  time 


FIG.  331 


unit,  then  the  altitude  pd  represents  the  acceleration,  at  the  same  scale 
at  which  the  velocity  ordinates  were  plotted.  In  other  words,  the 
slope  of  the  tangent  to  the  velocity-time  curve  represents  the  accel- 
eration. 


266 


ELEMENTS  OF  MECHANISM 


Crank 


FIG.  332 


F/'xed  Axis 
B 


Line  of  Centers 


Fixed  Axis 
G 


FIG.  333 


LINKWORK  267 

239.  Swinging  or  Rocking  Block  Four-Bar  Linkage.  Referring  to 
Fig.  321,  if,  instead  of  considering  the  piece  T  as  stationary,  giving  the 
four-bar  linkage  already  discussed,  the  piece  BC  is  made  stationary, 
the  linkage  becomes  that  shown  in  Fig.  332.  Here  BA  is  a  crank  but 
A  is  now  the  crank  pin.  The  pin  C  is  a  fixed  axis  with  the  block  swing- 
ing on  it.  As  the  crank  BA  revolves  the  piece  T  oscillates  and,  at  the 
same  time,  slides  back  and  forth  over  the  block. 

This  mechanism  is  applied  in  a  modified  form  as  a  quick  return 
motion  in  certain  machine  tools,  particularly  shapers.  Fig.  333  is  the 
same  as  Fig.  332,  drawn  with  the  line  BC  vertical  and  with  the  piece  T 
extended  with  a  connection  at  N  to  drive  the  ram  of  the  shaper.  The 
apparent  objection  to  using  the  mechanism  in  this  form  is  that  the  point 
N  moves  up  and  down  at];the  same  time  that  it  swings.  The  only  real 
purpose  of  the  sliding  of  the  piece  T  over  the  block  is  to  allow  the  dis- 
tance between  A  and  C  to  change  as  the  crank  BA  revolves.  This  is 
accomplished  equally  well  if  the  block  is  pivoted  on  the  pin  A  and  the 
arm  T  swings  on  a  pin  at  C,  as  shown  in  Fig.  334.  Fig.  335  shows 
one  way  in  which  the  mechanism  is  actually  used.  The  crank  BA  is 
in  this  case  the  gear  M.  A  is  a  pin  fast  to  M  and  carries  a  block 
which  works  in  a  slot  in  T.  M  is  driven  by  the  pinion  K.  The  link 
H  connects  N  to  the  ram  R  which  carries  the  cutting  tool.  The  length 
of  the  crank  BA  may  be  changed  by  turning  the  screw  S,  thus  changing 
the  length  of  stroke  of  the  tool. 

Ratio  of  Time  of  Cutting  Stroke  to  Time  of  Return  Stroke.  Fig.  336 
represents,  in  diagrammatic  form,  the  mechanism  of  Fig.  335.  The 
driving  crank  BA  turns  right-handed  as  shown  by  the  arrow.  When 
the  linkage  is  in  the  position  CNiRi  the  tool  slide  R  is  about  to  start 
on  the  cutting  stroke.  It  makes  this  stroke  while  CN  turns  through 
the  angle  NiCN  or  BA  turns  through  the  angle  a.  Similarly,  R  makes 
its  return  stroke  while  BA  is  turning  through  the  angle  /3.  If  BA  is 
assumed  to  turn  at  a  uniform  angular  speed  it  follows  that 

Time  required  for  cutting  stroke  _  a. 
Time  required  for  return  stroke  ~~  /3  ' 

Angular  Speed  of  Swinging  Arm.  The  angular  speed  of  the  swinging 
arm  CN  for  any  position  as  CNZ  may  be  found  by  the  principle  of 
§!228,  remembering  that  the  infinite  crank  at  C  is  always  perpendicular 
toCN. 

From  Equation  (73) 

Angular  speed  CNZ  _  Bm 
Angular  speed  BA  ~  CA2 


268 


ELEMENTS  OF  MECHANISM 


LINKWORK 


269 


Then  if  the  angular  speed  of  BA2  is  assumed  to  be  one  radian  per  unit 

of  time  gm 

Angular  speed  of  CNz  =  ~-T-  •  (85) 


Linear  Speed  of  N.  Since  the  linear  speed  of  a  point  on  a  swinging 
arm  is  equal  to  the  angular  speed  of  the  arm  multiplied  by  the  distance 
of  the  point  from  the  axis,  the  linear  speed  of  Nz  becomes,  from  Equa- 
tion (85), 


Linear  speed 


CA 


(86) 


\ 


Intersecting' 
atO 


/r._|/c 


\ 


(CiMting  Stroke 
(Return  Stroke 


Line  of  Centers 


.       FIG.  336 

Linear  Speed  of  Tool  Slide.  If  the  point  of  intersection  of  CN2  and 
the  perpendicular  to  RiR  at  R*  (which  in  this  case  falls  outside  the 
limits  of  the  drawing)  is  represented  by  the  letter  0,  then  0  is  the 
instantaneous  axis  of  N%Rz  and 

Linear  speed  of  #2  _  0#2. 
Linear  speed  of  N%      ON* 


(87) 


270 


ELEMENTS  OF   MECHANISM 


By  drawing  XXi  parallel  to  RzNz,  Equation  (87)  may  be  written 
Linear  speed  of  R2  _  KR2 
Linear  speed  of  N2      TN2 

Substituting  in  (88)  the  linear  speed  of  N2  obtained  in  (86)  gives 


(88) 


Linear  speed  of  R2  =  CN2  X  -       X  7 


(89) 

When,  as  is  usually  the  case,  RiR  is  perpendicular  to  CB,  the  triangle 
N2Cy,  formed  by  the  intersection  of  R2N2  (produced  if  necessary),  with 
CB  produced,  is  similar  to  triangle  R2ON2)  therefore 

0#2        Cy_ 

ON2      CN2 
Hence  Equation  (87)  may  be  written 

Linear  speed  of  R2  _  Cy  ,Q  , 

Linear  speed  of  N2~  CN~2 
or 

Linear  speed  of  R2  =  CN2  X  ^  -  X  -^r  =  -^  -  X  Cy.        (91) 


It  should  be  remembered  that  Equation  (89)  is  general  and  that  Equa- 
tion (91)  applies  only  when  the  path  of  the  tool  slide  is  perpendicular 
to  the  line  of  centers  CB.  From  each  of  the  Equations  (89)  and  (91)  the 
actual  linear  speed  of  the  tool  slide  is  obtained  by  multiplying  the  right- 
hand  side  of  the  equation  by  the  angular  speed  of  the  driving  crank, 
expressed  in  radians  per  unit  of  time. 


FIG.  337 

Oscillating  Engine.  Fig.  337  shows  in  diagrammatic  form  a  type  of 
engine,  known  as  an  oscillating  engine,  which  still  further  illustrates 
the  application  of  the  swinging  block  linkage.  Here  the  crank  pin  A 


LINKWORK 


271 


is  connected  directly  to  the  end  of  the  piston  rod,  and  turning  is  made 
possible  by  pivoting  the  cylinder  on  trunnions  at  C.  The  links  of  the 
equivalent  four-bar  linkage  are  indicated  in  the  figure. 

240.  Turning  Block  Linkage.  If  BA  (Fig.  321)  is  made  the  fixed 
link,  the  four-bar  linkage  becomes  that  shown  in  Fig.  338.  Here  BC  is 
a  crank  turning  about  B  and  causing  T  to  turn  about  A.  The  ratio  of 


Line  of 
Centers 


\ 


FIG.  338 

the  angular  speed  of  T  to  that  of  A  is  variable,  since  the  ratio  of  the 
perpendiculars  from  B  and  A  to  the  center  line  of  the  theoretical  con- 
necting rod  varies. 

Whitworth  Quick  Return  is  the  name  given  to  the  linkage  when  it  is 
used  as  a  quick-return  mechanism,  as  in  Fig.  339.  If  the  crank  BC 
(Fig.  339)  turns  uniformly  right-handed  from  the  position  C\  to  the 
position  €2,  the  slide  R  will  travel  from  its  extreme  position  at  the  right 
to  the  end  of  its  stroke  at  the  left;  and  while  BC  turns  from  C2  to  C\, 
the  slide  R  returns 

.    Time  of  cutting  stroke  of  R  _  a 
Time  of  return  stroke  of  R      (3 

To  locate  the  center  A,  given  the  time  ratio  of  cutting  stroke  to 
return  stroke,  the  line  of  centers,  the  axis  B  and  the  crank  BC;  make 


T     „  „  A  ,    .      8       ,         a       time  of  cutting;  stroke 

the  angle  CiBA  equal  to  £;>  where  -  =  TT—    —£ —  —  i  and 

2  |8        time  of  return  stroke 

draw  CiA  through  Ci  perpendicular  to  the  line  "of  centers;  the  point  A 


272 


ELEMENTS  OF   MECHANISM 


is  the  axis  of  the  link  AN.  If  the  stroke  of  the  slide  R  is  not  on  a  line 
passing  through  A,  but  below  it,  as  is  commonly  the  case,  the  time 
ratio  of  cutting  stroke  to  return  stroke  would  be  somewhat  different 
from  the  above. 

The  ratio  of  angular  speeds  of  AN  and  BC  and  the  linear  speed  of  R 
for  any  position  of  the  mechanism  may  be  found  by  a  method  corre- 


FIG.  339 

spending  to  that  used  for  the  swinging  block  quick  return,  described 
in  §  239. 

It  will  be  noticed  that  this  is  similar  in  appearance  to  the  swinging 
block  mechanism  shown  in  Figs.  333  and  334,  except  that  in  this  case 
the  driving  crank  is  longer  than  the  line  of  centers,  so  that  the  piece  T 


FIG.  340 


makes  a  complete  revolution  when  BC  turns  once,  whereas  the  arm  T 
in  Figs.  333  and  334  merely  oscillates. 

Fig.  340  shows  in  more  detail  the  manner  in  which  this  mechanism 
is  actually  applied. 


LINKWORK 


273 


241.  Sliding-Slot  Linkage.  If  the  block,  Fig.  321,  is  fixed  to  the 
frame  so  that  it  can  neither  turn  nor  slide,  the  resulting  mechanism  is  as 
indicated  in  Fig.  341.  The  link  CB  becomes  a  crank  oscillating  about 


Crank 


Connect/no 
Bod    ' 


FIG.  341 

the  fixed  axis  C.  The  connecting  rod  BA  may  make  complete  turns 
about  the  axis  A,  at  the  same  time  that  A  moves  in  a  straight  line, 
carrying  T  with  it.  If  BA  makes  a  complete  turn  relative  to  A  the 
stroke  of  T  is  equal  to  2  BA. 

Fig.  342  illustrates  a  manner  in  which  this  mechanism  may  be  applied. 
The  parts  are  lettered  to  correspond  with  Fig.  341.     The  worm  wheel 


FIG.  342 


carrying  the  pin  B  forms  the  connecting  rod.  This  wheel  may  be  made 
to  rotate  about  the  axis  A  by  a  worm  keyed  to  the  shaft  T.  The  worm 
and  wheel  are  kept  in  contact  by  a  piece  which  supports  the  bearing  of 
A,  hangs  from  the  shaft  T,  and  confines  the  worm  between  its  bearings. 


274 


ELEMENTS  OF  MECHANISM 


A  rotation  of  the  shaft  T  will  turn  the  worm  wheel,  causing  a  recipro- 
cation of  the  axis  A,  and  consequently  of  the  driving  shaft  T,  through  a 
distance  equal  to  twice  AB. 

Fig.  342a  is  another  application  of  this  linkage. 


Rod 


Crank 


FIG.  342a 

242.  Expansion  of  Elements  in  the  Linkages  with  One  Sliding 
Pair.  In  the  preceding  discussions  no  consideration  has  been  given 
to  the  diameters  of  the  cylindrical  pairs  of  elements  included  in  the 
linkages,  nor  to  the  size  and  form  of  the  links  themselves,  except  as 
they  were  touched  on  incidentally  in  illustrating  the  manner  of  apply- 
ing some  of  the  mechanisms,  as  in  Figs.  335,  340  and  342. 

Alterations  in  the  diameters  of  the  pairs  do  not  affect  the  relative 
motions.  Also  a  change  in  shape  or  size  of  the  links  does  not  alter  the 
relative  motions,  so  long  as  the  center  lines  of  the  elementary  links 
remain  unchanged,  and  yet  such  change  may  make  the  action  of  the 
linkage  possible.  Since  enlargements  of  the  elements  of  the  cylindric 
pairs  and  changes  in  the  form  of  the  links  sometimes  conceal  the  real 
nature  of  the  mechanism  and  cause  much  indistinctness,  it  will  be  well 
to  consider  a  few  cases  here. 

The  sliding-block  linkage  of  Fig.  332  will  first  be  considered.  Each 
of  its  links  is  more  or  less  closely  connected  with  its  three  cylindric 
pairs,  and  their  forms  are  therefore  dependent  upon  the  relative  sizes 
of  the  latter,  although  this  size  does  not  affect  the  nature  of  their  rela- 
tive motion.  Evidently  the  combination  is  not  altered  kinematically, 
if  the  diameter  of  the  crank  shaft  is  increased  so  as  to  include  the  crank 


LINKWORK 


275 


pin  as  shown  in  Fig.  343,  where  the  different  links  are  lettered  the  same 
as  in  Fig.  332.  S  is  the  enlarged  shaft  whose  center  is  A.  The  open 
cylinder  of  the  fixed  piece  T  must  be  enlarged  to  the  same  extent  as 
the  shaft,  so  that  the  pair  is  still  closed. 

This  arrangement  is  used  in  practice,  in  some  slotting  and  shearing 
machines,  to  work  a  short-stroke  pump  from  the  end   of  an  engine 


M 


FIG.  343 

shaft;  and  in  other  cases  where  a  short  crank  forms  one  piece  with  its 
own  shaft. 

If  the  crank  pin  is  enlarged  until  it  includes  the  shaft,  the  result  is 
the  common  eccentric  and  eccentric  rod  shown  in  Fig.  344.     Here 


FIG.  344 

the  eccentric  is  E  and  is  in  reality  a  circular  disk  with  center  at  B, 
made  fast  to  the  crank  shaft  so  that  it  revolves  with  the  shaft  about 
the  axis  A.  The  length  of  the  equivalent  crank  AB  is  called  the 
eccentricity.  This  mechanism,  which  differs  in  form  only  from  the 
common  crank  and  connecting  rod,  is  much  used  to  operate  the  valve 
mechanism  in  steam  engines,  where  it  is  necessary  to  obtain  a  recip- 
rocating motion,  often  less  than  the  diameter  of  the  engine  shaft.  The 
part  of  the  rod  K  which  encloses  E  is  called  the  eccentric  strap,  and  is 
made  in  two  parts,  and  may  be  separate  from  Ky  the  eccentric  rod,  which  is 
usually  bolted  to  one  of  these  parts;  the  cylindrical  pair  is  also  so  shaped 
as  to  allow  no  axial  motion  of  K  on  E. 


276 


ELEMENTS  OF  MECHANISM 


If  the  crank  pin  is  still  further  expanded  until  it  includes  the  cross- 
head  pin,  the  arrangement  shown  in  Fig.  345  is  obtained.  In  this  case, 
the  element  of  the  cylindric  pair  which  belongs  to  the  crank  AB  has 


FIG.  345 

been  inverted,  and  thus  made  open.  The  rod  EC  becomes  an  eccentric 
disk  which  swings  about  the  axis  of  bearing  in  the  piece  M,  and  is  always 
in  contact  with  the  hollow  disk  E,  carried  by  the  shaft  turning  in  T. 


FIG.  346 


If,  instead  of  enlarging  the  crank  pin  to  include  the  crosshead  pin, 
the  latter  is  enlarged  to  include  the  former,  the  arrangement  shown  in 
Fig.  346  is  obtained.  The  rod  BC  is  again  an  eccentric  disk  or  annular 


LINKWORK 


277 


ring;  but  it  now  oscillates  in  a  ring  forming  part  of  the  piece  M,  while 
the  crank  pin  drives  it  by  internal  contact.  In  order  to  make  the 
relations  of  these  expansions  more  clear,  fine  light  lines  have  been 
drawn  in  each  case,  showing  the  elementary  links.  The  above  exhausts 
all  the  practicable  combinations  of  the  three  cylindric  pairs. 

In  Fig.  346  the  link  K  may  be  replaced  by  an  annular  ring  containing 
the  crank  pin  and  oscillating  in  a  corresponding  annular  groove  in  the 
piece  M.  So  long  as  the  center  of  this  ring  remains  the  same  as  that  of 
K,  the  mechanism  remains  unchanged,  and  as  the  motion  of  the  ring 
is  merely  oscillatory,  only  a  sector  of  it  need  be  used  and  enough  of  the 
annular  groove  to  admit  of  sufficient  motion  of  the  sector  in  its  swing. 
Fig.  347  represents  the  arrangement  altered  in  this  way,  the  different 


M 


FIG.  347 

parts  being  lettered  the  same  as  in  Fig.  346,  BC  is  still  the  connecting 

rod  replaced  by  K  and  its  motion  as  a  link  in  the  chain  remains  the 

same  as  before,  and  is  completely  restrained;   the  shape  of  the  sector 

always  fixes  the  length  of  the  connecting  rod. 

This  mechanism  is  made  use  of  in  the  Stevenson 

and  Gooch  reversing  gears  for  locomotives,  and 

in  other  places;  the  chains  are  not  there  simple, 

but  compound.     It  will  be  noticed  that  if  the 

radius  of  the  slot  is  made  infinite  the  result  will 

be  as  in  Fig.  325. 

The  mechanism  shown  in  Fig.  348,  which  some- 
times occurs  in  slotting  and  metal-punching 
machines,  is  another  illustration  of  pin  expansion. 
The  whole  forms  a  sliding-block  linkage;  the 
link  K  is  formed  essentially  as  in  Fig.  347,  but 
here  the  profiles  against  which  it  works  are 
concave  on  both  sides  of  the  crank  pin,  the  upper 


FIG.  348 


profile  being  of  large,  and  the  lower  of  very  small,  radius,  but  both  forming 
part  of  the  block  M.    The  work  is  done  when  the  block  M  is  moving 


278  ELEMENTS  OF  MECHANISM 

downwards,  and  the  small  radius  profile  being  then  in  use,  the  friction 
is  reduced.  In  this  case  the  block  M  is  so  enclosed  by  the  guides  T 
that  the  profiles  representing  the  crosshead  pin  lie  entirely  within  the 
sliding  pair,  an  illustration  of  how  the  method  of  expansion  can  be 
applied  to  the  fourth  or  sliding  pair. 

The  illustrations  in  Figs.  335,  340  and  342  furnish  further  examples 
of  the  expansion  or  modification  of  various  parts  of  the  linkage  in 
order  to  make  the  mechanical  application  possible. 

A  change  in  the  shape  of  an  elementary  link  frequently  permits 
motions  to  take  place  which  are  not  otherwise  possible.  In  Fig.  349, 


FIG.  349 

for  example,  a  complete  rotation  of  BA  to  cause  a  reciprocation  of  M 
would  be  possible  with  the  open  rod  K  moving  around  the  fixed  shaft 
E,  but  not  with  the  elementary  link  BC,  shown  by  a  light  line. 

243.  The  Isosceles  Linkage.  If  the  link  BC  (Fig.  321)  is  made 
equal  in  length  to  A  B  the  result  is  a  series  of  mechanisms  analagous  to 
those  obtained  from  Fig.  321  by  fixing  successively  the  four  links. 
The  mechanisms  obtained,  however,  have  special  properties  due  to  the 
equality  of  A  B  and  BC.  Moreover,  it  will  be  seen  that  the  four  cases 
obtained  by  fixing  the  four  links  are  reduced  to  two  since  the  same 
result,  kinematically,  is  obtained  by  fixing  the  block  M  as  by  fixing 
the  piece  T  and  the  axis  A ;  and  by  fixing  BC  as  by  fixing  AB. 

The  Isosceles  Sliding-Block  Linkage.  If  the  piece  T  is  fixed  as  in 
Fig.  350  the  mechanism  corresponds  to  that  of  Fig.  322.  Here  (Fig. 
350),  if  AB  is  the  driver  and  C  starts  from  the  position  Ci  it  will  be 
found  when  the  crank  A  B  is  at  an  angle  of  90°  with  ACi  (the  path 
of  C)  that  C  is  directly  over  A  and  any  further  rotation  of  A  B  will 
cause  only  a  similar  rotation  of  CB.  In  order  to  cause  C  to  continue 
in  its  path  from  this  position  it  will  be  necessary  to  pair  points  on  the 
centrode  of  BC  for  this  position  (when  T  is  fixed)  with  the  correspond- 
ing points  on  the  centrode  of  T  (when  BC  is  fixed)  as  was  done  in  the 
case  of  the  anti-parallel  crank  linkage  in  Figs.  316  and  318. 

The  centrode  of  BC  is  the  circle  drawn  about  A  as  a  center  with 
radius  2  AB.  This  can  be  seen  as  follows : 


LINKWORK 


279 


In  any  position  of  the  linkage,  as  that  occupied  in  Fig.  350,  produce 
A  B  to  meet  the  perpendicular  to  ACi,  through  C,  at  0,  thus  finding  the 
instantaneous  axis  for  that  position.  From  B  draw  Bk  perpendicular 

AC 
to  ACi]  then,  since  ABC  is  an  isosceles  triangle,  Ak  =  kC  =  -75—    Hence, 

A 


from  the  similarity  of  the  triangles  AOC  and  ABk,  AB 


AO 

--  •    This  holds 


m 

for  every  position  of  the  linkage.    Therefore  the  locus  of  0,  the  instan- 
taneous axis  of  BC,  is  the  circle  with  radius  2  AB. 

A  similar  method  of  reasoning  can  be  followed  to  show  that  the  cen- 
trode  of  T7,  with  BC  fixed,  is  a  circle  about  B  with  radius  BA. 


*cmn  acting  Rod 


FIG.  350 

From  the  properties  of  centrodes  previously  brought  out,  if  the  link 
BC  is  made  fast  to  a  disk  of  radius  AB  ( =  BC)  with  the  point  B  at  its 
center  and  C  on  its  circumference,  and  this  disk  is  rolled  inside  a  fixed 
hollow  cylinder  of  twice  its  own  diameter,  BC  will  have  the  same 
motion  that  it  would  if  it  were  the  connecting  rod  of  the  actual  four-bar 
linkage  ABC,  and  C  will  travel  on  a  diameter  of  the  larger  circle. 

It  is  evident  that  since  BC  is  a  radius  of  the  centrode  of  T  (that  is, 
of  the  disk  just  referred  to)  and  C  has  a  motion  along  the  diameter 
CiAC2,  if  BC  is  prolonged  to  E,  making  BE  equal  to  BC,  E  will  travel 
the  diameter  E3Ei}  being  at  A  when  C  is  at  d,  at  E3  when  AB  and  BC 


ELEMENTS  OF  MECHANISM 


280 


are  perpendicular  to  C^ACi  above  CiAC2,  at  A  again  when  C  is  at  C2, 
and  at  E±  when  AB  is  perpendicularly  under  CiAC2. 

If  now,  when  the  actual  linkage  is  used,  with  the  connecting  rod 
prolonged  to  E,  a  pin  is  centered  at  E  and  corresponding  fixed  eyes 
at  Es  and  E*  as  in  Fig.  351,  a  means  is  provided  for  causing  C  to 
continue  in  its  path  when  AB  is  in  the  90°  positions. 


FIG.  351 

It  should  be  noticed  that  the  paths  of  the  points  C  and  E,  Fig.  350, 
as  shown  above,  when  considered  as  points  of  the  circumference  of  the 
smaller  circle  (that  is,  on  the  surface  of  the  centrode  of  T)  are  hypo- 
cycloids  with  the  circle  AO  as  the  directing  circle  and  circle  BO  as  the 
generating  circle.  From  this  it  is  evident  that  the  prolongation  BE 
need  not  be  in  the  same  line  as  BC  but  may  be  at  any  angle  as  at  BE6, 
provided  the  eye  is  properly  located. 

If  the  crank  AB  turns  at  uniform  angular  speed  C  has  harmonic 
motion  over  the  path  CiAC2  for  C  is  always  found  at  the  foot  of  the 
perpendicular  OC  and  0  is  always  on  the  line  AB  produced  distant 
2AB  from  0.  This  agrees  with  the  description  given  for  harmonic 
motion  in  Chapter  I. 

In  Fig.  352,  lettered  to  correspond  to  Fig.  350,  the  actual  crank  AB  is 
omitted  and  a  block  is  placed  at  the  end  of  the  rod  BE  to  guide  E  in  a 
slot  whose  center  line  is  a  straight  line  passing  through  A.  The  nature 
of  the  linkage  remains  the  same  as  in  Fig.  350.  The  imaginary  line 
joining  B,  the  middle  point  of  EC,  to  A,  the  point  of  intersection  of  the 
center  lines  of  the  slots  is  still  the  theoretical  crank  and,  if  motion  is 


LINKWORK 


281 


imparted  to  C,  B  will  move  in  a  circular  path  about  A.  The  whole 
may  be  thought  of  as  two  four-bar  linkages  ABCD  and  ABEDi,  with 
the  crank  AB  common  to  the  two  linkages. 

The  Elliptic  Trammel,  which  is  so  commonly  used  for  drawing  ellipses, 
is  an  application  of  the  principle  of  the  isosceles  sliding-block  linkage. 
Referring  to  the  actual  linkage  as  in  Fig.  353,  and  its  equivalent  as  in 
Fig.  354  (with  CBE  a  straight  line),  it  has  been  shown  that  E  (Fig.  353) 
moves  in  a  straight  line  through  A  and  that  B  (Fig.  354)  moves  in  a 


Connecting  Rod 


Lme  of  Centers 


^K 


D  at  infinity 


FIG.  352 

circle  about  A.  If  any  other  point,  as  P,  on  the  rod  CE  or  CE  pro- 
longed, is  chosen,  P  can  be  shown  to  move  in  a  path  which  is  an  ellipse 
with  axes  lying  along  the  paths  of  C  and  E  and  with  semi-axes  equal  in 
length  to  PE  and  PC.  If  PE  is  less  than  PC  the  minor  axis  lies  along 
the  path  of  C.  If  PC  is  less  than  PE  the  minor  axis  lies  along  the  path 
of  E,  as  in  the  figure. 

In  the  elliptic  trammel  the  mechanism  is  usually  applied  in  the  form 
corresponding  to  Fig.  354  and  the  ellipse  is  usually  traced  by  an  ad- 
justable point  outside  of  E  or  C  as  in  the  figure;  E  and  C  are  made  so 
that  their  distance  apart  is  adjustable  and  they  are  set  one-half  the 
difference  of  the  major  and  minor  axes  apart. 

An  ellipse  can  be  readily  drawn  by  taking  a  card  one  corner  of  which 
shall  represent  the  tracing  point  P.  Points  corresponding  to  the  de- 


282 


ELEMENTS  OF  MECHANISM 


sired  positions  of  E  and  C  are  then  marked  on  the  edge  of  the  card, 
and  by  placing  these  points  in  successive  positions  on  lines  at  right 
angles  with  each  other,  corresponding  to  the  slots  in  which  the  blocks 
in  Fig.  354  move,  and  marking  the  successive  positions  of  P,  will  give 
a  series  of  points  on  the  required  ellipse. 


FIG.  353 

To  prove  that  the  point  P  moves  on  an  ellipse,  let  Pn  =  x;  Pr  =  y; 
PE  (semi-major  axis)  =  a;  PC  (semi-minor  axis)  =  6. 
The  equation  of  an  ellipse  referred  to  the  center  as  the  origin  is 

a?  +  V==  l' 
In  Figs.  353  and  354 

x      Pn 
a      PE 

and,  since  the  triangles  nPE  and  rCP  are  similar, 

Pr  _  nE 
PC~  PE' 

Pn    ,    Pr       Pn    ,  nE 
Therefore  ™  +  -^  =  ™  +  -^ 

JL  Jli  ±\j  JL  Hi  r  TJ 

,   .     ...  Pn2      Pr2      Pn2    .  nE2       Pn2  -f  nE2 

and,  in  this  case,     =r  +  =^  =  =-  +  =r  =  — =- —  =  1. 
PE2      PC2      PE2      PE2  PE2 

Therefore  \  +  ^  =  1, 

showing  that  the  locus  of  P  is  an  ellipse. 

By  fixing  the  link  BC,  Fig.  350,  or  the  equivalent,  fixing  the  centers 
C  and  E  (allowing  the  blocks  to  turn),  Fig.  352  or  Fig.  354,  the  mechan- 
ism corresponding  to  the  swinging-block  linkage,  Fig.  332,  is  obtained. 


LINKWORK 


283 


Two  examples  will  be  considered  in  which  this  linkage  is  expanded  in 
the  manner  suggested  in  Figs.  352  and  354. 

The  Elliptic  Chuck  depends  upon  the  principle  proved  for  the  elliptic 
trammel  and  upon  the  principle,  previously  referred  to,  that  the  relative 
motions  of  the  parts  of  a  linkage  are  independent  of  the  fixedness  of 
the  links. 

Now  in  drawing  an  ellipse  with  a  trammel,  the  paper  is  fixed,  and 
the  pencil  is  moved  over  it;  but  in  turning  an  ellipse  in  a  lathe,  the 
tool,  which  has  the  same  position  as  the  pencil,  is  fixed,  and  the  piece  to 
be  turned  should  have  such  a  motion  as  would  compel  the  tool  to  cut 


ellipses.  This  is  accomplished  in  the  elliptic  chuck,  in  which  the 
spindle  of  the  lathe,  with  a  block  on  it,  corresponds  to  the  axis  E  of 
Fig.  354.  In  another  fixed  bearing  whose  axis  corresponds  to  C  is 
another  shaft  having  a  block  on  it.  The  point  of  the  cutting  tool  is  in 
a  fixed  position  corresponding  to  P. 

The  piece  carrying  the  work  corresponds  to  T  and  has  two  slots 
at  right  angles,  sliding  over  the  blocks  on  the  spindle  and  the  axis  C. 
The  turning  of  the  spindle  causes  the  point  of  intersection  of  the  center 
lines  of  these  slots  to  move  in  a  circle  about  the  axis  of  the  spindle 
and  the  whole  piece  to  have  such  a  motion  that  the  point  of  the  tool 
cuts  an  ellipse  from  the  material  attached  to  it. 


284 


ELEMENTS  OF  MECHANISM 


The  Oldhams  Coupling  shown  in  Fig.  355  is  an  interesting  example  of 
this  form  of  linkage.  The  axis  E  of  the  upper  shaft  corresponds  to  E 
in  Fig.  354  and  the  slotted  disk  on  this  shaft  corresponds  to  the  block 
at  E\  similarly,  with  the  shaft  at  C.  The  intermediate  disk  T  with 
two  projections  at  right  angles  across  its  diameters  replaces  the  cross  T. 


FIG.  355 

The  object  of  the  device  is  to  connect  two  parallel  shafts  placed  a 
short  distance  apart  so  as  to  communicate  uniform  rotation  from  one 
to  the  other. 

If  the  link  A B,  Fig.  350,  is  made  the  stationary  link  the  result  is  the 
same,  kinematically,  as  the  linkage  discussed  in  the  last  two  illustra- 


FIG.  356 

tions.    The  details  of  application  are  somewhat  different, 
shows  the  mechanism. 


Fig.  356 


LINKWORK 


285 


FIG.  357 


If  T  is  on  a  shaft  centered  at  A  the  crank  BC,  through  the  arm  T, 
will  cause  the  shaft  at  A  to  turn  at  an  angular  speed  equal  to  one-half 
its  own.  There  may  be  two  arms  BC  and  BE  with  corresponding 
slots,  the  result  being  the  equivalent  of  a  two-toothed  wheel  driving, 
internally,  one  of  four  teeth.  If  there  were  three  arms  and  three  slots 
the  equivalent  gears  would  have  three  and  six  teeth. 

244.  Other  Forms  of  Linkwork  with  Two  Sliding  Pairs.  Fig.  357 
shows  a  combination  of  an  eccentric  circular 
disk  A,  and  a  sliding  piece  C,  moving  through 
fixed  guides,  one  of  which  is  shown  at  D.  A 
uniform  rotation  of  A  about  the  axis  a  will 
give  harmonic  motion  to  C.  This  can  be 
shown  by  noticing  that  the  distance  which  C 
has  moved  from  its  highest  position  is 

cd  =  ef  =  ab  (1  -  cos  6), 

which  is  the   equation  for  simple  harmonic 

motion  where  2  ab  is  the  stroke  of  the  slide. 
This  mechanism  can  also  be  found  by  an 

expansion  of  the  crank  pin  B,  Fig.  325,  until  it 

includes  the  shaft  A}  the  slot  being  correspondingly  enlarged,  ana  then 

after  turning  the  figure  through  90°,  omitting  the  lower  part  of  the 

cross,  allowing  the  pairing  to  be  by  force-closure. 

The  Swash-Plate.    The  apparatus  shown  in  Fig.  358,  known  as  a 

swash-plate,  consists  of  an  elliptical  plate  A  set  obliquely  upon  the 
shaft  8,  which  by  its  rotation  causes  a  sliding  bar  C  to 
move  up  and  down,  in  a  line  parallel  to  the  axis  of  the 
shaft,  in  the  guides  D,  the  friction  between  the  end  of 
the  bar  and  the  plate  being  lessened  by  a  small  roller  0. 
When  a  roller  is  used,  the  motion  of  the  bar  C  is  approxi- 
mately harmonic  —  the  smaller  the  roller  the  closer  the 
approximation.  If  a  point  is  used  in  place  of  the 
roller,  the  motion  is  harmonic,  which  can  be  shown  as 
follows : 

Since  the  bar  C  remains  always  parallel  to  the  axis  of 
the  shaft,  the  path  of  the  point  0,  projected  upon  an 

imaginary  plane  through  the  lowest  position  of  0  and  perpendicular 

to  the  shaft  8,  will  be  a  circle,  and  the  actual  path  of  0  on  the  plate 

A  will  be  an  ellipse. 

In  Fig.  359  let  eba  represent  the  angular  inclination  of  the  plate  to 

the  axis  of  the  shaft,  ab  the  axis  of  the  shaft,  eof  the  actual  path  of  the 

point  o  on  the  plate,  and  the  dotted  circle  erd  the  projection  of  this 


FIG.  358 


286 


ELEMENTS  OF  MECHANISM 


path  upon  a  plane  through  e  (the  lowest  position  of  o)  perpendicular 
to  the  axis  ab. 

Draw  om  perpendicular  to  ef,  or  perpendicular  to  the  plane  erd,  and 
rn  perpendicular  to  ed,  the  diameter  of  the  circle  erd.  Join  mn,  and 
suppose  the  plate  to  rotate  through  an  angle  ear  =  6,  and  thus  to  carry 
the  point  o  through  a  vertical  distance  equal  to  or. 

Then 


,  v   en  f     mn      en\ 

or  =  mn  =  ab  X  —  [  as  — r  =  — ) 

ea  \      ab       ea/ 

(ea  —  an\ 
ea     ) 


FIG.  359 


=  ab 

=  ab[l-  — 
\        &aj 

=  ab(l  —  cos  6), 


or  the  same  formula  as  was  derived  in  the  case  of  harmonic  motion.  In 
this  case  ab  represents  the  length  of  the  equivalent  crank,  and  is  equal 
in  length  to  one-half  of  the  stroke  of  the  rod  C. 

245.  The  Conic  Four-Bar  Linkage.  If  the  axes  of  the  four  cylin- 
dric  pairs  of  the  four-bar  linkage  are  not  parallel,  but  have  a  common 
point  of  intersection  at  a  finite  distance,  the  chain  remains  movable 
and  also  closed  (Fig.  360).  The  lengths  of  the  different  links  will  now 
be  measured  on  the  surface  of  a  sphere 
whose  center  is  at  the  point  of  inter- 
section of  the  axes.  The  axoids  will  no 
longer  be  cylinders,  but  cones,  as  all  the 
instantaneous  axes  must  pass  through 
the  common  point  of  intersection  of  the 
pin  axes. 

The  different  forms  of  the  cylindric 
linkage  repeat  themselves  in  the  conical 
one,  but  with  certain  differences  in  their 
relations.  The  principal  difference  is  in  the  relative  lengths  of  the  links, 
which  would  vary  if  they  were  measured  upon  spherical  surfaces  of 
different  radii,  the  links  being  necessarily  located  at  different  distances 
from  the  center  of  the  sphere  in  order  that  they  may  pass  each  other  in 
their  motions.  The  ratio,  however,  between  the  length  of  a  link  and  its 
radius  remains  constant  for  all  values  of  the  radius,  and  these  ratios  are 
merely  the  values  of  the  circular  measures  of  the  angles  subtended  by 
the  links.  In  place  of  the  link  lengths,  the  relative  magnitudes  of  these 
angles  can  be  considered. 


FIG.  360 


LINKWORK 


287 


The  alterations  in  the  lengths  of  the  links  will  now  be  represented  by 
corresponding  angular  changes.  The  infinitely  long  link  corresponds 
to  an  angle  of  90°,  as  this  gives  motion  on  a  great  circle  which  corre- 
sponds to  straight-line  motion  in  the  cylindric  linkages. 

Fig.  361  shows  plan  and  elevation  of  a  conic  four-bar  linkage  ABCD, 
the  link  AB  turning  about  A,  and,  for  a  complete  turn,  causing  an 
oscillation  of  the  link  CD  about  D  through  the  angle  0,  shown  in  the 
elevation.  In  the  figure  each  of  the  links  BC  and  CD  subtends  90°, 
while  the  link  AB  subtends  about  30°.  Varying  the  angles  which  the 

links  subtend  will,  of  course,  vary  the  relative 
motions  of  AB  and  CD. 

246.  Hooke's  Joint.  If  in  Fig.  361  each 
of  the  links  AB,  BC,  and  CD  is  made  to  sub- 
tend an  angle  of  90°,  AB  and  CD  will  each 
\G  make  complete  rotations.  This  mechanism, 
known  as  a  Hooke's  joint,  is  represented  by 
Fig.  362,  A  and  D  are  the  two  intersecting 
shafts,  and  the  links  AB  and  CD,  fast  to  the 
shafts  A  and  D  re- 
spectively, subtend 
90°,  while  the  con- 
necting link  BC  also 
subtends  90°. 

In  order  to  make 
the  apparatus 
stronger  and  stiff er, 
two  sets  of  links  are 
used,  and  the  link 
CB  is  continued 
around  as  shown, 
thus  giving  an 
annular  ring  joining  the  ends  of  the  double  links  CDC'  and  BAB'.  This 
ring  is  sometimes  replaced  by  a  sphere  into  which  the  pins  C,  B,  Cr, 
and  B'  are  fitted,  or  by  a  rectangular  cross  with  arms  of  a  circular 
section  working  in  the  circular  holes  at  B,  C,  C',  and  B'.  Or,  the  arms 
BABi  and  CACi  may  be  paired  with  grooves  cut  in  a  sphere  in  planes 
passing  through  the  center  of  the  sphere  and  at  right  angles  to  each 
other.  Such  forms  of  Hooke's  joint  are  much  used. 

Relative  Motion  of  the  Two  Connected  Shafts.  —  Given  the  angular 
motion  of  AB,  to  find  the  angle  through  which  CD  turns.  Fig.  363 
shows  a  plan  and  elevation  of  a  Hooke's  joint,  so  drawn  that  the  axis 
A  is  perpendicular  to  the  plane  of  elevation.  If  the  link  AB  is  turned 


C2 


FIG.  361 


FIG.  362 


288 


ELEMENTS   OF  MECHANISM 


through  an  angle  6,  it  will  be  projected  in  the  position  ABi.  The  path 
of  the  point  C  will  be  on  a  great  circle  in  a  plane  perpendicular  to  the 
axis  D,  which  will  appear  in  the  elevation  as  the  ellipse  BCE.  The 

point  C  will  then  move  to  C1}  found  by  making 
the  angle  B^ACi  equal  to  90°,  for  the  link  BC 
subtends  90°,  and  since  the  radius  from  B  to 
the  center  of  the  sphere  is  always  parallel  to 
the  plane  of  elevation,  its  projection  and  that 
of  the  radius  from  C  will  always  be  at  right 
angles.  The  projected  position  of  the  linkage 
after  turning  A  through  the  angle  B  will  be 
ABiCiD.  To  find  the  true  angle  through 
which  the  link  CD  and  the  shaft  D  have 
turned,  swing  the  ellipse  BCE  with  the  axis 
D,  until  D  is  perpendicular  to  the  plane  of 
elevation,  when  the  points  C  and  Ci  will  be 
found  at  C'  and  d',  respectively,  giving  the 
angle  C\ACr  =  </>  as  the  true  angle  through 
which  the  axis  D  has  turned.  Or  the  arm 
DCi  may  be  revolved  until  shaft  D  is  perpen- 
dicular to  the  horizontal  plane,  giving  CiBCi 
=  0,  as  shown  in  the  plan. 

It  is  evident  from  the  above  that  two 
intersecting  shafts  connected  by  a  single 
Hooke's  joint  cannot  have  uniform  motions. 
If,  however,  two  joints  are  used  to  connect 
two  parallel  or  intersecting  shafts,  they  may  be  so  arranged  that  they 
will  have  uniform  motions. 

247.  Double  Hooke's  Joint.     Two  parallel  or  intersecting  shafts  may 
be  connected  by  a  double  Hooke's  joint  and  have  uniform  motions,  provided 
that  the  intermediate  shaft  makes  equal  angles  with  the  connected  shafts, 
and  that  the  links  on  the  intermediate  shaft  are  in  the  same  plane.     Fig. 
364  gives  a  plan  and  elevation  of  two  shafts  so  connected,  and  the  posi- 
tion after  turning  through  an  angle  d.     It  is  evident  that  one  joint  just 
neutralizes  the  effect  of  the  other. 

The  term  universal  joint  is  often  used  to  designate  the  above- 
described  mechanism. 

248.  Angular  Speed  Ratio  in  a  Single  Hooke's  Joint.     Fig.  365 
reproduces  the  elevation  given  in  Fig.  363,  which  shows  the  angles 
6  and  0  through  which  A  and  D  move  respectively.     The  angle  EAC 
=  a  will  be  the  true  angle  between  the  planes  in  which  the  paths  of  the 
points  B  and  C  lie;  to  find  the  angle  <j>  analytically  in  terms  of  d  and  a, 
we  have,  from  Fig.  365, 


LINKWORK 


289 


=  CVG      §£        tan0  =  ^£ 


~AG      C1 

.'.    tan  0  =  tan  6  cos  a. 


_  _  AC  = 

AG      AC'      AE~ 


(92) 


To  obtain  the  velocity  ratio,  differentiate  equation  (92),  remembering 
that  cos  a  is  a  constant;  then 

* '       sec2  0  1  +  tan2  0  x™ 

-  cos  a.  (y6) 


FIG.  364      . 

If  we  eliminate  <£  and  0  from  equation  (93),  by  use  of  equation  (92) 
we  shall  obtain 

d<f>  cos  a 

d0  ~  1  —  sin2  0  sin2  a 

_  1  —  cos2  0  sin2  a 
cos  a 


(95) 


290 


ELEMENTS  OF  MECHANISM 


Assume  AB  and  CD  the  starting  positions  of  the  arms  AB  and  CD 
respectively;  then  equations  (94)  and  (95)  will  have  minimum  values 
when  sin  6  =  0  and  cos  0  =  1;  this  will  happen  when  6  and  0  are  0°  and 

180°,  giving  -^  =  cos  a  in  both  cases.    Thus  the  minimum  velocity  ratio 
(W 

occurs  when  the  driving  arm  is  at  AB  and  AB2,  the  corresponding  posi- 
tions of  the  following  arm  being  CD  and 
C%D.  Maximum  values  occur  when  sin  0 

s\     ji        dd>  1 

0;  then-^ 


=  1  and  cos 


which 


dO       cos  a 

will  happen  when  0  and  0  are  90°  and  270°, 
the  corresponding  positions  of  the  driving 
arm  being  ABs  and  AB*. 

Hence  in  one  rotation  of  the  driving 
shaft  the  velocity  ratio  varies  twice  be- 

and  cos  a;  and  be- 


tween  the  limits 


cos  a 

tween  these  points  there  are  four  positions 
where  the  value  is  unity. 

If  the  angle  a  increases,  the  variation  in  the  angular  velocity  ratio  of 
the  two  connected  shafts  also  increases;  and  when  this  variation  becomes 
too  great  to  be  admissible  in  any  case,  other  arrangements  must  be 
employed. 


CHAPTER  XII 
STRAIGHT-LINE  MECHANISMS  —  PARALLEL  MOTIONS 

249.  A  Straight-line  Mechanism  is  a  linkage  designed  to  guide  a 
reciprocating  piece  either  exactly  or  approximately  in  a  straight  line, 
in  order  to  avoid  the  friction  arising  from  the  use  of  straight  guides. 
Some  straight-line  mechanisms  are  exact,  that  is,  they  guide  the  recipro- 
cating piece  in  an  exact  straight  line;    others,  which  occur  more  fre- 
quently, are  approximate,  and  are  usually  designed  so  that  the  middle 
and  two  extreme  positions  of  the  guided  point  shall  be  in  one  straight  line, 
while  at  the  same  time  care  is  taken  that  the  intermediate  positions 
deviate  as  little  as  possible  from  that  line. 

250.  Peaucellier's   Straight-line   Mechanism.     Fig.   366   shows  a 
linkage,  invented  by  M.  Peaucellier,  for  describing  an  exact  straight 
line  within  the  limits  of  its  motion. 

It  consists  of  eight  links  joined  at  their  ends.     Four  of  these  links, 
A,  B,  C,  and  D}  are  equal  to  each  other  and  form  a  cell;  the  two  equal 


FIG.  366 

links  E  and  F  connect  the  opposite  points  of  the  cell,  a  and  e,  with  the 
fixed  center  of  motion  d;  the  link  G  =  %  bd  oscillates  on  the  fixed  center 
c,  cd  thus  forming  the  fixed  link  equal  in  length  to  G. 

If  now  the  linkage  be  moved  within  the  working  limits  of  its  con- 
struction (that  is,  until  the  links  B  and  G,  and  C  and  G  come  into  line 
on  opposite  sides  of  the  center  line  of  motion  cd),  the  cell  will  open  and 

291 


292  ELEMENTS  OF  MECHANISM 

close;  the  points  a  and  e  will  describe  circular  arcs  about  d,  and  b  about  c. 
Finally,  the  point  p  will  describe  a  straight  line  ss  perpendicular  to  the 
line  of  centers  cd. 

To  prove  this,  move  the  linkage  into  some  other  position,  as  piaibicd. 
(It  is  to  be  noticed  that  since  the  links  A  and  D,  B  and  C,  and  E  and  F 
always  form  isosceles  triangles  with  a  common  base,  a  straight  line  from 
p  to  d  will  always  pass  through  b.)  If  the  line  traced  by  the  point  p  is 
a  straight  line,  the  angle  pipd  will  be  90°.  The  angle  bbid  is  90°,  since 
be  =  cd  =  bic;  therefore  the  triangles  p\pd  and  bbid  would  be  similar 
right  triangles,  and  we  should  have 

pd  _  bid 
pid      bd 

To  prove  that  ss  is  a  straight  line  it  is  necessary  to  show  that  the 
above  relation  exists  in  the  different  positions  of  the  linkage.  In  Fig.  366 

E2  =  tf 2  +  (bf  +  bd)2-, 

£2  =  ?2  +  6/2; 
...    #2  _  #2  =  2  (bf)  (bd)  +  bd2  =  bd  (bd  +  2  bf). 

But,  since  the  links  A  and  B  are  equal,  the  triangle  pah  is  isosceles  and 
the  base  pb  =  2  bf. 

.-.    E*-B2=(bd)(pd).  (96) 

By  the  same  process,  when  the  linkage  is  in  any -other  position,  as 
Placed,  we  should  have 

Ef  -  Bi*  =  (bid)  (pid).  (97) 

Equating  equations  (44)  and  (45), 

(bd)  (pd)  =  (hd)  (pid), 

pd       bid 

or  ^}  =  TT> 

pid       bd 

which  proves  that  the  path  of  the  point  p  is  on  the  straight  line  ss. 

If  the  relation  between  the  links  cd  and  be  be  taken  different  from 
that  shown  (Fig.  366),  the  points  b  and  p,  sometimes  called  the  poles 
of  the  cell,  will  be  found  to  describe  circular  arcs  whose  centers  are  on 
the  line  passing  through  c  and  d;  in  the  case  shown,  one  of  these  cir- 
cular arcs  has  a  radius  infinity. 

251.  Scott  Russell's  Straight-line  Mechanism.  This  mechanism, 
suggested  by  Mr.  Scott  Russell,  is  an  application  of  the  isosceles  sliding- 
block  linkage. 

It  is  made  up  of  the  links  ab  and  pc,  Fig.  367.  The  link  ab,  centered 
at  a,  is  joined  to  the  middle  point  b  of  the  link  pc,  and  ab,  be  and  pb 
are  taken  equal  to  each  other;  and  the  point  c  is  constrained  to  move 


STRAIGHT-LINE  MECHANISMS 


293 


in  the  straight  line  ac  by  means  of  the  sliding  block.  In  this  case  the 
motion  of  the  sliding  block  c  is  slight,  as  the  entire  motion  of  p  is  seldom 
taken  as  great  as  cp. 

To  show  that  the  point  p  describes  a  straight  line  ppipz  perpendicular 
to  ac  through  a,  a  semicircle  may  be  drawn  through  p  and  c  with  b  as  a 
center;  it  will  also  pass  through  a  so  that  pac  will  be  a  right  angle- 
therefore  the  point  p  is  on  apt 
which  is  true  for  all  positions 
of  p. 

The  point  a  should  be  located 
in  the  middle  of  the  path  or 
stroke  of  p.  The  motion  of  c 


may  then 
equation 


be   found   by  the 


FIG.  367 


where  ap  is  the  half -stroke  of  p. 

Approximate    straight-line 
motions  somewhat  resembling 
the  preceding  may  be  obtained  by  guiding  the  link  cp  entirely  by 
oscillating  links,  instead  of  by  a  link  and  slide. 

1°  In  the  link  cp  (Fig.  367)  choose  a  convenient  point  e  whose  mean 
position  is  ei,  and  whose  extreme  positions  are  e  and  e2.  Through  these 
three  points  pass  a  circular  arc,  ee^,  the  center  of  which  /  will  be  found 
on  the  line  ac.  Join  e  and  /  by  a  link  ef,  and  the  two  links  ab  and  ef 
will  so  guide  pe  that  the  mean  and  extreme  positions  of  p  will  be  found 
on  the  line  pp2,  provided  suitable  pairs  are  supplied  to  cause  passage  by 
the  central  position. 

2°  The  point  c  may  be  made  to  move  very  nearly  in  a  straight  line 
cci  by  means  of  a  link  cd  centered  on  a  perpendicular  erected  at  the 
middle  point  of  the  path  of  cf  The  longer  this  link  the  nearer  the  path 
of  c  will  approach  a  straight  line. 

This  straight-line  motion  has  been  applied  in  a  form  of  small  sta- 
tionary engines,  commonly  known  as  grasshopper  engines,  where  cbp 
(Fig.  367) ,  extended  beyond  p,  forms  the  beam  of  the  engine,  its  right- 
hand  end  being  supported  by  the  link  cd.  The  piston-rod  is  attached, 
by  means  of  a  crosshead,  to  the  point  p,  which  describes  a  straight  line, 
and  the  connecting-rod  is  attached  to  a  point  in  the  line  cp  produced, 
both  piston-rod  and  connecting-rod  passing  downward  from  cp.  In 
this  case  it  will  be  noticed  that  the  pressure  on  the  fulcrum  c,  of  the 
beam,  is  equal  to  the  difference  of  the  pressures  on  the  crosshead  pin 
and  crank-pin  instead  of  the  sum,  as  in  the  ordinary  form  of  beam  engine. 


294 


ELEMENTS  OF  MECHANISM 


In  this  second  form  of  motion  it  is  not  always  convenient  to  place  the 
point  a  in  the  line  of  motion  pp2,  and  it  is  often  located  on  one  side,  as 
shown  in  Fig.  368 

P 


FIG.  368 


The  proportions  of  the  different  links  which  will  cause  the  point  p 
to  be  nearly  on  the  straight  line  at  the  extreme  positions  and  at  the 
middle  may  be  found  as  follows: 

Let  pg  be  one-half  the  stroke  of  the  point  p,  and  let  the  angle  bac  =  0, 
and  bca  =  pbe  =  <£.     In  this  extreme  position  we  may  write 
ag  =  of  -  fg  =  af  -  be 
=  ab  cos  0  —  pb  cos  <£ 


=  ab  (l  -  2  sin2|)  -  pb(l  -  2sin2|Y 


But  if  the  links  are  taken  long  enough,  so  that  for  a  given  stroke  the 
angles  6  and  0  are  small,  then  sin  6  =  6,  nearly,  and  sin  <£  —  <£,  nearly, 
and 


-4-;0-4-J) 

=  ab  —  pb  —  ab  -^  +  pb  -^  • 


or 


If  the  linkage  is  now  placed  in  its  mid-position, 

ag  =  ab  —  pb. 
Equating  Equations  (98)  and  (99), 

,   6*        ,02 

ab2=pb~2' 

ab  _tf 

pb~  fl2' 


But  in  the  triangle  abc 


ab 
be 
ab 


0 


sn 


sin  6      6 


nearly; 


2 
be 


or 


(98) 


(99) 


(100) 


(101) 


STRAIGHT-LINE  MECHANISMS  295 

Hence  the  links  must  be  so  proportioned  that  be  is  a  mean  propor- 
tional between  ab  and  pb,  which  also  holds  true  when  the  path  of  p 
falls  to  the  left  of  a  instead  of  between  a  and  c. 

As  an  example  of  the  case  where  the  path  of  the  guided  point  falls 
to  the  left  of  a  we  have  the  straight-line  motion  of  the  Thompson 
steam-engine  indicator,  Fig.  374. 

252.  Watt's  "Straight-line  Mechanism.  Fig.  369  shows  a  Watt 
straight-line  mechanism.  Here  the  two  links  ad  and  be  connected  by 
the  link  ab  oscillate  on  the  fixed 
centers  d  and  c,  and  any  point,  as 
p,  in  the  connecting  link  ab  will 
describe  a  complex  curve.  If  the 
point  p  be  properly  chosen,  a 
double-looped  curve  will  be  ob- 
tained, two  parts  of  which  are 
nearly  straight  lines.  In  designing 
such  a  motion  it  is  customary  to 
use  only  a  portion  ef  of  one  of  the 
approximate  straight  lines,  and  to  Fi 

so  proportion  the  different  links 
that  the  extreme  and  middle  points  e,  f,  and  p  shall  be  on  a  line  per- 
pendicular to  the  center  lines  of  the  levers  ad  and  be  in  their  middle 
positions,  when  they  should  be  taken  parallel  to  each  other. 

The  linkage  is  shown  in  its  mid-position  by  dabc,  Fig.  370,  and  in  the 
upper  extreme  position  by  daibic,  where  ppi  is  to  be  one-half  the  stroke 
of  p.  Given  the  positions  of  the  links  ad  and  be  when  in  their  mid- 
position,  the  axes  c  and  d,  the  line  of  stroke  ss,  and  the  length  of  the 
stroke  desired;  to  find  the  points  a  ancffr,  giving  the  link  ab,  and  to 
prove  that  the  point  p,  where  ab  crosses  ss,  will  be  found  on  the  line  ss 
when  it  is  moved  up  (or  down)  one-half  the  given  stroke.  Lay  off 
on  ss  from  the  points  g  and  h,  where  the  links  ad  and  be  cross  the  line 
ss,  one-quarter  of  the  stroke,  giving  the  points  k  and  l\  connect  these 
points  with  the  axes  d  and  c  respectively;  draw  the  lines  akai  and  blbi 
perpendicular  to  dk  and  el  respectively,  making  aai  =  2ak  and  661  =  2  bl; 
then  if  the  link  centered  at  d  were  ad,  it  could  swing  to  aid,  and  similarly 
be  could  swing  to  bic.  By  construction  kg  =  \  stroke,  and  aai  =  2ak; 
therefore  a^e  =  |  stroke.  Similarly  bif  =  \  stroke,  which  would  make 
the  figure  eaibif  a  parallelogram,  and  ai6i  would  equal  ef.  But  ef  is 
equal  to  ab,  since  bh  =  hf  and  ag  =  ge.  Therefore,  if  the  linkage  is 
dbac,  it  can  occupy  the  position  daibic',  and  since  ap  =  ep  =  aipi,  and 
ppi  =  eai  =  J  the  stroke,  the  point  d  will  be  at  pi  and  J  the  stroke 
above  p. 


296 


ELEMENTS  OF  MECHANISM 


To  calculate  the  lengths  of  the  links,  given  dg,  ch,  and  gh,  and  the 
length  of  stroke  S.  Since  the  chord  aai  of  the  arc  through  which  a 
would  move  is  bisected  at  right  angles  by  the  line  dk, 

S2 
gk2  =  to)  (dg)  =  — 


Similarly, 


ag  = 


be  =  ch  + 


and 

S2   . 


FIG.  370 

To  determine  the  position  of  the  point  p  we  have  from  the  figure 

ap  :  bp  =  ag  :  bh.  (102) 


from  which 
or 


ap:bp=l6dg:!6ch  =  C 

ap  :  ab  =  ch  :  ch  +  dg, 
bp  :  ab  =  dg  :  ch  +  dg, 


STRAIGHT-LINE  MECHANISMS 


297 


from  which  the  position  of  the  guided  point  p  can  be  calculated.     If,  as 
is  very  often  the  case,  ad  =  be,  then 

S2 
ad  =  bc==dg+—, 

and  bp  :  ab  =  dg  :  2  dg, 

or  bp  =  |  ab, 

and  the  point  p  is  thus  at  the  middle  of  the  link  ab. 

This  mechanism  may  be  arranged  as  shown  in  Fig.  371,  where  the 
centers  c  and  d  are  on  the  same  side  of  the  line  of  motion.     The  graphical 


FIG.  371 

solution  is  the  same  as  in  Fig.  370,  with  the  result  that  p  is  found  where 
ab  extended  crosses  the  line  of  stroke  ss,  and,  as  before,  it  can  be  shown 
that  if  p  is  moved  up  one-half  the  given  stroke,  it  will  be  found  on  the 
line  of  stroke  ss. 

In  Fig.  370,  letting  the  angle  ada\  =  6  and  bcbi  =  <£,  we  have,  from 
Equation  (102), 

_  2  Q 

j/i  /i\      ad  sin2:r 

ap_ag_ae_ad(l—  cos  6)  __  2 

~  ~ 


bp      bh      bf  ~  6c  (1  —  cos 


r-2   •  2  > 
oc  sin  _ 


298 


ELEMENTS  OF  MECHANISM 


which  may  be  written 


==  _  v 
~ 


sin2?- 


Bin'* 


But  ad  sin  6  =  be  sin  <£;  and  since  the  angles  6  or  <f>  would  rarely  exceed 
20°,  we  may  assume  that 


ad  sin  •=  =  be  sin  -• 

2>  JL 

ap       be  , 

•  •     t    =  ~5 '  nearly, 
bp      ad 


(103) 


or  the  segments  of  the  link  are  inversely  proportional  to  the  lengths  of 
the  nearer  levers,  which  is  the  rule  usually  employed  when  the  extreme 
positions  can  vary  a  very  little  from  the  straight  line.  When  the  levers 
are  equal  this  rule  is  exact. 

253.  The  Pantograph.  The  pantograph  is  a  four-bar  linkage  so 
arranged  as  to  form  a  parallelogram  abed,  Fig.  372.  Fixing  some  point 
in  the  linkage,  as  e,  certain  other  points,  as/,  g,  and  h,  will  move  parallel 
and  similar  to  each  other  over  any  path  either  straight  or  curved.  These 
points,  as  f,  g,  and  h,  must  lie  on  the  same  straight  line  passing  through 

f 


FIG.  372 

the  fixed  point  e,  and  their  motions  will  then  be  proportional  to  their  distances 
from  the  fixed  point.  To  prove  that  this  is  so,  move  the  point  /  to  any 
other  position,  as/i;  the  linkage  will  then  be  found  to  occupy  the  posi- 
tion aibiCidi.  Connect /i  with  e',  then  hi,  where  fie  crosses  the  link  bid, 
can  be  proved  to  be  the  same  distance  from  c\  that  h  is  from  c,  and  the 
line  hhi  will  be  parallel  to  ffi. 


STRAIGHT-LINE  MECHANISMS  299 

In  the  original  position,  since  fd  is  parallel  to  he,  we  may  write 

fd_de_  fe 
he      ce      he 

In  the  second  position,  since  fidi  is  parallel  to  h\c\  and  since  fie  is  drawn 
a  straight  line,  we  have 


Now  in  these  equations  —  =  —  J    therefore  %-  =  T—  J    but  /d  =  /idi, 

ce      Cie  /ic      hid 

which  gives  he  —  hid,  which  proves  that  the  point  h  has  moved  to  hi. 

Also  7^  =  T^->  from  which  it  follows  that  ffi  is  parallel  to  Mi,  and 
he      hie 

ffL=fe  =  dej 
hhi      he      ce 

or  the  motions  are  proportional  to  the  distances  of  the  points  /  and  h 
from  e. 

To  connect  two  points,  as  a  and  b,  Fig.  373,  by  a  pantograph,  so  that 
their  motions  shall  be  parallel  and  similar  and  in  a  given  ratio,  we  have, 
first,  that  the  fixed  point  c  must  be  on  the  straight  line  ab  continued, 


FIG.  373 

and  so  located  that  ac  is  to  be  as  the  desired  ratio  of  the  motion  of  a  to  b. 
After  locating  c,  an  infinite  number  of  pantographs  might  be  drawn. 
Care  must  be  taken  that  the  links  are  so  proportioned  as  to  allow  the 
desired  magnitude  and  direction  of  motion. 

It  is  interesting  to  note  that  if  b  were  the  fixed  point,  a  and  c  would 
move  in  opposite  directions.  It  can  be  shown  as  before  that  their 
motions  would  be  parallel  and  as  ab  is  to  be. 

The  pantograph  is  often  used  to  reduce  or  enlarge  drawings,  for  it 
is  evident  that  similar  curves  may  be  traced  as  well  as  straight  lines. 
Also  pantographs  are  used  to  increase  or  reduce  motion  in  some  definite 
proportion,  as  in  the  indicator  rig  on  an  engine  where  the  motion  of 
the  crosshead  is  reduced  proportionally  to  the  desired  length  of  the 
indicator  diagram.  When  the  points,  as/ and  h  (Fig.  372),  are  required 
to  move  in  parallel  straight  lines  it  is  not  always  necessary  to  employ 
a  complete  parallelogram,  provided  the  mechanism  is  such  that  the 


300 


ELEMENTS  OF  MECHANISM 


points  /  and  h  are  properly  guided.  Such  a  case  is  shown  in  Fig.  374, 
which  is  a  diagram  of  the  mechanism  for  moving  the  pencil  on  a  Thomp- 
son steam-engine  indicator.  The  pencil  at  /,  which  traces  the  diagram 
on  a  paper  carried  by  an  oscillating  drum,  is  guided  by  a  Scott-Russell 
straight-line  motion  abed  so  that  it  moves  nearly  in  a  straight  line  ss 
parallel  to  the  axis  of  the  drum,  and  to  the  center  line  of  the  cylinder 
it.  It  must  also  be  arranged  that  the  motion  of  the  pencil  /  always 
bears  the  same  relation  to  the  motion  of  the  piston  of  the  indicator  on 
the  line  it.  To  secure  this  draw  a  line  from  /  to  d  and  note  the  point  e 
where  it  crosses  the  line  it.  Then  e  will  be  a  point  on  the  piston-rod,  which 
rod  is  guided  in  an  exact  straight  line  by  the  cylinder.  If  now  the  link 
eh  is  added  so  that  its  center  line  is  parallel  to  cd,  we  should  have, 


FIG.  374 

assuming  /  to  move  on  an  exact  straight  line,  the  motion  of  /  parallel  to 
the  motion  of  e  and  in  a  constant  ratio  as  cf  :  ch  or  as.  df  :  de.  This  can 
be  seen  by  supposing  the  link  eg  to  be  added,  which  completes  the 
pantograph  dgehcf.  If  eg  were  added,  the  link  ab  could  not  be  used,  as 
the  linkage  abcdf  does  not  give  an  exact  straight-line  motion  to  /.  For 
constructive  reasons  the  link  eg  is  omitted;  a  ball  joint  is  located  at  e 
which  moves  in  an  exact  straight  line,  and  the  point  /  is  guided  by  the 
Scott-Russell  motion,  the  error  in  the  motion  being  very  slight  indeed. 
Slides  are  often  substituted,  in  the  manner  just  explained,  for  links 
of  a  pantograph,  and  exact  reductions  are  thereby  obtained.  In  Fig. 
375  the  points  /  and  h  are  made  to  move  on  the  parallel  lines  mm  and  nn 


STRAIGHT-LINE  MECHANISMS 


301 


respectively.  Suppose  it  is  desired  to  have  the  point  h  move  J-as 
much  as  /.  Draw  the  line  fhe  and  lay  off  the  point  e  so  that  eh  :  ef 
=  1:3;  draw  a  line,  as  ed,  and  locate  a  point  d  upon  it  which  when 
connected  to  /  with  a  link  df  will  move 
nearly  an  equal  distance  to  the  right 
and  left  of  the  line  ef  and  above  and 
below  the  line  mm  for  the  known  motion 
of  /.  Draw  ch  through  h  and  parallel 
to  df.  The  linkage  echdf  will  accomplish 
the  result  required.  The  dotted  link  ah 
may  be  added  to  complete  the  panto- 
graph, and  the  slide  h  may  then  be 
removed  or  not  as  desired.  The  figure 
also  shows  how  a  point  g  may  be  made 
to  move  in  the  opposite  direction  to  / 
in  the  same  ratio  as  h  but  on  the  line 
nini,  the  equivalent  pantograph  being  drawn  dotted, 
shown  in  its  extreme  position  to  the  left  by  heavy  lines  and  to  the  right 
by  light  lines.  9 

254.  Combination  of  Watt's  Straight-line  Mechanism  with  a  Pan- 
tograph. Watt's  straight-line  mechanism  has  been  much  used  in 
beam  engines,  and  it  is  generally  necessary  to  arrange  so  that  more 
than  one  point  can  be  guided,  which  is  accomplished  by  a  pantograph 
attachment. 

In  Fig.  376  a  straight-line  mechanism  is  arranged  to  guide  three 
points  p,  pi,  and  p%  in  parallel  straight  lines.  The  case  chosen  is  that 


375 

The  link  ed  is 


FIG.  376 

of  a  compound  condensing  beam  engine,  where  PZ  is  the  piston  rod  of 
the  low-pressure  cylinder,  PI  that  of  the  high-pressure  cylinder,  and  P 


302  ELEMENTS  OF  MECHANISM 

the  pump  rod,  all  of  which  should  move  in  parallel  straight  lines,  per- 
pendicular to  the  center  line  of  the  beam  in  its  middle  position. 

The  fundamental  linkage  dabc  is  arranged  to  guide  the  point  p  as 
required;  then  adding  the  parallelograms  asib  and  ap^rb,  placing  the 
links  st  and  pzr  so  that  they  pass  through  the  points  p\  and  p2  respect- 
ively, found  by  drawing  the  straight  line  cp  and  noting  points  pi  and 
pz  where  it  intersects  lines  PI  and  PZ,  we  obtain  the  complete  linkage. 
The  links  are  arranged  in  two  sets,  and  the  rods  are  carried  between 
them;  the  links  da  are  also  placed  outside  of  the  links  p^a.  When  the 
point  p  falls  within  the  beam  a  double  pump-rod  must  be  used.  The 
linkage  is  shown  in  its  extreme  upper  position  to  render  its  construction 
clearer. 

The  various  links  are  usually  designated  as  follows:  cr  the  main 
beam,  ad  the  radius  bar  or  bridle,  p%r  the  main  link,  ab  the  back  link, 
and  p%a  the  parallel  bar,  connecting  the  main  and  back  links. 

In  order  to  proportion  the  linkage  so  that  the  point  p%  shall  fall  at 
the  end  of  the  link  rpz  we  have,  by  similar  triangles  cbp  and  crpz, 

cb  :bp  =  cr  :  rpz  =  cr  :  ab. 

ab  X  cb     - 

/.    cr  =  — 7 

op 

The  relative  stroke  S  of  the  point  p2  and  s  of  the  point  p  are  expressed 
by  the  equation 

S  :  s  =  cpz :  cp  =  cr  :  cb. 

If  we  denote  by  M  and  N  the  lengths  of  the  perpendiculars  dropped 
from  c  to  the  lines  of  motion  P2  and  P  respectively,  then 

S:s  =  M  :N 

and  S  =  8W'    S  =  Slti'  (104) 

The  problem  will  generally  be,  given  the  centers  of  the  main  beam 
c  and  bridle  d,  the  stroke  S  of  the  point  pz,  and  the  paths  of  the  guided 
points  p,  pi,  and  pz,  to  find  the  remaining  parts.  The  strokes  of  the 
guided  points  can  be  found  from  Equation  (104)  and  then  the  method  of 
§  252,  Fig.  370,  can  be  applied. 

255.  Robert's  Approximate  Straight-line  Mechanism.  This  might 
also  be  called  the  W  straight-line  mechanism,  and  is  shown  in  Fig.  377. 
It  consists  of  a  rigid  triangular  frame  abp  forming  an  isosceles  triangle 
on  ab,  the  points  a  and  b  being  guided  by  links  ad  =  be  =  bp,  oscillat- 
ing on  the  centers  d  and  c  respectively,  which  are  on  the  line  of  motion  dc. 

To  lay  out  the  motion,  let  dc  be  the  straight  line  of  the  stroke  along 
which  the  guided  point  p  is  to  move  approximately,  and  p  be  the  mid- 


STRAIGHT-LINE  MECHANISM 


303 


FIG.  377 


ac 


die  point  of  that  line.  Draw  two  equal  isosceles  triangles,  dap  and  cbp; 
join  ab,  which  must  equal  dp  =  pc.  Then  abp  is  the  rigid  triangular 
frame,  p  the  guided  point,  and  d  and  c  are 
the  centers  of  the  two  links.  The  extreme 
positions  when  p  is  at  d  and  c  are  shown  at 
daiaz  and  cajh,  the  point  02  being  common  to 
both.  The  length  of  each  side  of  the  triangle, 
as  ap  =  da,  should  not  be  less  than  1.186  dp, 
since  in  this  case  the  points  cc^i  and  dovb2  lie 
in  straight  lines.  It  may  be  made  as  much 
greater  as  the  space  will  permit,  and  the  greater  it  is  the  more  accurate 
will  the  motion  be.  The  intermediate  positions  between  dp  and  cp  vary- 
somewhat  from  the  line  dc. 

256.  Tchebicheff's   Approximate    Straight-line   Mechanism.     Fig. 
378  shows  another  mechanism  giving  a  close  approximation  to  a  straight- 
line  motion  invented  by  Prof.  Tchebicheff. 

The  links  are  made  in  the  following  proportion:  If  cd  =  4,  then 
bd  =  5  and  ab  =  2.  The  guided  point  p  is  located  midway  be- 
tween a  and  b  on  the  link  ab  and  is  distant  from  cd  an 
amount  equal  to  \/52  —  32  =  4.  When  the  point  p 
moves  to  pi,  directly  over  d,  dpi  =  dbi  —  bipi  =  5  —  1 
=  4.  Thus  the  middle  and  extreme  positions  of  p,  as 
shown,  are  in  line,  but  the  intermediate  positions  will 
be  found  to  deviate  slightly  from  the  straight  line. 
To  render  the  range  of  motion  shown  on  the  figure 
possible  the  links  ac  and  bd  would  need  to  be  offset. 

257.  Parallel  Motion  by  Means  of  Four-bar  Linkage.     The  parallel 
crank  mechanism,  §  234,  Fig.  313,  is  very  often  used  to  produce  parallel 
motions.     The  common  parallel  ruler,  consisting  of  two  parallel  straight- 
edges connected  by  two  equal  and  parallel  links  is  a  familiar  example 
of  such  application.     A  double  parallel  crank  mechanism  is  applied  in 
the  Universal  Drafting-machine,  extensively  used  in  place  of  T  square 
and  triangles.     Its  essential  features  are  shown  in  Fig.  379.    The  clamp 
C  is  made  fast  to  the  upper  left-hand  edge  of  the  drawing-board  and 
supports  the  first  linkage  abed.     The  ring  cedf  carries  the  second  linkage 
efhg,  guiding  the  head  P.    The  two  combined  scales  and  straight- 
edges A  and  B,  fixed  at  right  angles  to  each  other,  are  arranged  to 
swivel  on  P,  and  by  means  of  a  graduated  circle  and  clamp-nut  may 
be  set  at  any  desired  angle,  the  device  thus  serving  as  .a  protractor. 
The  fine  lines  show  how  the  linkages  appear  when  the  head  is  moved 
to  Pi,  and  it  is  easily  seen  that  the  straight-edges  will  always  be  guided 
into  parallel  positions. 


FIG.  378 


304 


ELEMENTS  OF  MECHANISM 


FIG.  379 


FIG.  380 


FIG.  381 


STRAIGHT-LINE  MECHANISMS 


305 


258.  Parallel  Motion  by  Cords.  Cords,  wire  ropes,  or  small  steel 
wires  are  frequently  used  to  compel  the  motion  of  long  narrow  car- 
riages or  sliders  into  parallel  positions.  In  Fig.  380  the  slider  R  has  at 
either  end  the  double-grooved  wheels  E  and  F.  A  cord  attached  to  the 
hook  A  passes  vertically  downward  under  F,  across  over  E,  and  down- 
ward to  the  hook  C.  A  similarly  arranged  cord  starts  from  B,  passes 
around  E  and  F,  using  the  remaining  grooves,  and  is  made  fast  to 
hook  D.  On  moving  the  slider  downward  it  will  be  seen  that  for  a 
motion  of  I"  the  wheel  F  will  give  out  I"  of  the  rope  from  A  and  take 
up  I"  of  rope  from  D,  which  is  only  possible  when  E  takes  up  I"  from 
C  and  gives  out  I"  to  B.  Thus  the 
slider  R  is  constrained  to  move  into 
parallel  positions.  In  practice  turn- 
buckles  or  other  means  are  provided  to 
keep  the  cords  taut. 

Figs.  381  and  382  show  two  other 
arrangements  which  will  accomplish  the 
same  purpose.  In  Fig.  381,  sometimes 
applied  to  guide  straight-edges  on 
drawing-boards,  the  cords  or  wires  cross 
on  the  back  side  of  the  board  where 
the  four  guide-wheels  are  located  and 
the  straight-edge  R  is  guided  by  special 
fastenings  E  and  F,  passing  around  the 
edges  of  and  under  the  board.  By 
making  one  of  these  fastenings  movable  the  straight-edge  may  be  ad- 
justed. Fig.  382  shows  a  similar  device  that  might  be  applied  on  a 
drawing-board.  Here  the  wires  are  on  the  front  of  the  board  and  are 
arranged  to  pass  under  the  straight-edge  in  a  suitable  groove.  The 
turnbuckle  T  serves  to  keep  the  wires  taut,  and  the  slotted  link  S  allows 
adjustment. 

The  device  shown  in  Fig.  380  is  often  known  as  a  squaring  band  and 
is  applied  in  spinning-mules  and  in  some  forms  of  travelling  cranes. 


FIG.  382 


CHAPTER  XIII 


MISCELLANEOUS  MECHANISMS  —  AGGREGATE  COMBINA- 
TIONS —  PULLEY  BLOCKS  —  INTERMITTENT  MOTION 

269.  Aggregate  Combinations  is  a  term  applied  to  assemblages  of 
pieces  in  mechanism  in  which  the  motion  of  the  follower  is  the  result- 
ant of  the  motions  given  to  it  by  more  than  one  driver.  The  number  of 
independently-acting  drivers  which  give  motion  to  the  follower  is  gen- 
erally two,  and  cannot  be  greater  than  three,  as  each  driver  determines 
the  motion  of  at  least  one  point  of  the  follower,  and  the  motion  of  three 
points  in  a  body  fixes  its  motion. 

By  means  of  aggregate  combinations  we  may  produce  very  rapid  or 
slow  movements  and  complex  paths,  which  could  not  well  be  obtained 
from  a  single  driver. 

The  epicyclic  gear  trains  discussed  in  Chapter  VII  in  reality  come 
under  the  heading  of  aggregate  combinations. 

260.  Aggregate  Motion  by  Linkwork.  Figs.  383  and  384  repre- 
sent the  usual  arrangement  of  such  a  combination.  A  rigid  bar  ab  has 

two  points,  as  a  and  6,  each  con- 
Q3  nected  with  one  driver,  while  c 
may  be  connected  with  a  follower. 
Let  aai  represent  the  linear  velocity 
of  a,  and  661  the  linear  velocity  of 
b:  to  find  the  linear  velocity  of  c. 
Consider  the  motions  to  take  place 
separately;  then  if  6  were  fixed, 
the  linear  velocity  aa\  given  to  a 
»  would  cause  c  to  have  a  velocity 

'          represented  by  cci.     Considering  a 
as  fixed,  the  linear  velocity  661  at  b 
would  give  to  c  a  velocity  ccz.     The 
FIG.  383  FIG.  384  aggregate  of  these  two  would  be 

the  algebraic  sum  of  cc\  and  ccz. 

In  Fig.  383  we  have  cci  acting  to  the  left,  while  cc^  acts  to  the  right; 
therefore  the  resulting  linear  velocity  of  c  will  be  cca  =  cci  —  ccz  acting 

306 


\  \  / 
\  \  / 
\  \  / 


\ 


MISCELLANEOUS   MECHANISMS 


307 


to  the  left,  since  cci  >  cc2.  In  Fig.  384,  where  both  cci  and  CC2  act  to 
the  left,  the  result  is  cc3  =  cci  +  cc2  acting  to  the  left.  It  will  be  seen 
that  the  same  results  could  have  been  obtained  by  finding  the  instan- 
taneous axis  o  of  ab  in  each  case,  when  we  should  have  linear  velocity 
c  :  linear  velocity  a  =  co  :  ao. 

In  many  cases  the  lines  of  motion  are  not  exactly  perpendicular 
to  the  link,  nor  parallel  to  each  other,  neither  do  the  points  a,  b,  and  c 
necessarily  lie  in  the  same  straight  line,  but  often  the  conditions  are 
approximately  as  assumed  in  Figs.  383  and  384,  so  that  the  error  intro- 
duced by  so  considering  them  may  be  sufficiently  small  to  be  practically 
disregarded. 

As  examples  of  aggregate  motion  by  linkwork  we  have  the  different 
forms  of  link  motions  as  used  in  the  valve  gears  of  reversing  steam 
engines.  Here  the  ends  of  the  links  are  driven  by  eccentrics,  and  the 
motion  for  the  valve  is  taken  from  some  intermediate  point  on  the  link 
whose  distance  from  the  ends  may  be  varied  at  will,  the  nearer  end 
having  proportionally  the  greater  influence  on  the  resulting  motion. 

A  wheel  rolling  upon  a  plane  is  a  case  of  aggregate  motion, 
the  center  of  the  wheel  moving  parallel  to  the  plane,  and 
the  wheel  itself  rotating  upon  its  center.  The  resultant  of 
these  two  motions  gives  the  aggregate  result  of  rolling. 

261.  Pulley-blocks  for 
Hoisting.  The  simple 
forms  of  hoisting-tackle, 
as  in  Fig.  385,  are  exam- 
ples of  aggregate  combi- 
nations. The  sheaves  C 
and  D  turn  on  a  fixed  axis, 
while  A  and  B  turn  on  a 
bearing  from  which  the 
weight  W  is  suspended. 
Fig.  386  is  in  effect  the 
same  as  Fig.  385,  but  gives 
a  clearer  diagram  for 
studying  the  linear  velo- 
city ratio.  Assume  that 

Fi  385  ^ne  kar  °h  wft-h  the  sheaves 
A  and  B  and  the  weight 
W  has  an  upward  velocity  represented  by  v. 
sheave  A,  since  the  point  c  at  any  instant  is  fixed,  is  equivalent 
to  a  wheel  rolling  on  a  plane,  and  there  would  be  an  upward  linear 
velocity  at  d  =  2  v.  At  the  sheave  B  there  is  the  aggregate  motion  due 


4V 


*7 


2v 


W 


FIG.  386 
The  effect  of  this  at  the 


308 


ELEMENTS  OF   MECHANISM 


rp,       ,. 
Therefore 


to  the  downward  linear  velocity  at  e  =  2  v  and  the  upward  linear  velocity 
of  the  axis  b  =  v,  giving  for  the  linear  velocity  of  /,  4  v  upwards. 
linear  speed  F  _  4  _  TF 
linear  speed  W  ~  I  ~  F 
Many  elevator-hoisting  mechanisms  are  arranged  in  a  similar  man- 
ner, the  force  being  applied  at  TF,  and  the  resulting  force  being  given  at  F. 
This  means  a  large  force  acting  through  a  relatively  small  distance, 
producing  a  relatively  small   force   acting  through  a   much   greater 
distance. 

The  mechanical  advantage  of  a  hoist  is  the  ratio  of  the  weight 
which  can  be  lifted  to  the  force  which  is  exerted,  friction  being 
neglected. 

The  mechanical  advantage  of  a  given  hoist  can  be  determined  by 
rinding  the  velocity  ratio  as  above  and  then,  since  the  distances  moved 
through  in  a  given  time  (assuming  constant  velocity  ratio)  are  directly 
as  the  velocities,  the  forces  must  be  inversely  as  the  velocities.  Other 
methods  of  determining  the  mechanical 
advantage  are  illustrated  by  the  fol- 
lowing examples.* 

Example  73.  Hoist  with  Two  Single  Sheave 
Blocks.  In  Fig.  387  the  upper  block  A, 
known  as  the  standing  block,  is  suspended 
from  a  fixed  support.  The  rope  is  made  fast 
to  the  casing  of  the  upper  block,  passes  around 
the  sheave  in  the  lower  block  and  up  around 
the  sheave  P  which  turns  about  the  axis  S  in 
the  upper  block.  It  is  required  to  find  the 
force  at  F  necessary  to  raise  a  weight  W  of  100 
Ibs.  suspended  from  the  lower  block. 

Solution.  Assume  that  W  is  lifted  1  ft.  by 
some  external  force  with  the  rope  at  F  not 
moving.  Then  1  ft.  of  slack  rope  would  result 
at  R  and  another  foot  of  slack  at  T,  giving  a 
total  of  2  ft.  of  slack  which  must  be  drawn 
over  to  F  in  order  to  keep  the  rope  tight. 
Therefore,  the  linear  speed  of  F  is  to  the  linear 
speed  of  W  as  2  is  to  1.  Hence,  F  is  to  W  as 
1  is  to  2,  or  F  =  |  W  =  50  Ibs. 

Example  74.  Hoist  with  One  Single  Block 
and  One  Double  Block.  The  hoist  shown  in 
Fig.  388  has  the  part  of  the  rope  which  is 
marked  T  made  fast  to  the  lower  block;  it 


-B 


W 


FIG.  387 


W 


FIG.  388 


then  passes  over  a  sheave  in  the  upper  block,  comes  down  at  R  and  passes  under  the 
sheave  in  the  lower  block,  then  up  at  P  over  a  second  sheave  in  the  upper  block 
and  off  at  F. 

*  These  solutions  assume  that  the  ropes  are  parallel. 


MISCELLANEOUS  MECHANISMS 


309 


It  is  required  to  find  the  mechanical  advantage  of  this  hoist;  that  is,  the  ratio 
of  the  weight  W  to  the  force  at  F. 

Solution.  Applying  the  same  method  used  in  Example  47,  shows  1  ft.  of  slack  in 
each  of  the  three  parts  R,  T,  and  P,  or  a  total  of  3  ft.  which  must  be  drawn  off  at 
F  if  W  is  lifted  1  ft.  by  an  external  force.  Therefore, 


W 
F 


Example  75.  "Luff  on  Luff"  Fig.  389  shows  a  combination  of  two  sets  of 
pulley  blocks,  the  rope  F  of  the  first  set  being  made  fast  to  the  moving  block  of  the 
second  set. 

Solution.  The  mechanical  advantage  of  each  set  is  found  as  in  the  previous  ex- 
amples. Then  the  product  of  the  two  is  the  mechanical  advantage  of  the  combina- 
tion. The  first  set  in  this  case  has  a  mechanical 
advantage  of  3,  and  the  second  set  of  4;  therefore 
the  combination  has  a  mechanical  advantage  of  12. 
If  the  hook  H  were  attached  to  a  stationary  support 
and  the  load  applied  to  the  hook  K,  the  advantage 
of  the  system  would  be  16. 


FIG.  389 


FIG.  390 


Example  76.  "Spanish  Burton."  If  the  weight  W  (Fig.  390)  is  lifted  1  ft.,  a 
foot  of  slack  is  caused  at  both  P  and  R.  The  foot  at  P  is  carried  over  to  T  which, 
in  turn,  causes  a  foot  of  slack  in  both  R  and  F;  this  makes  a  total  of  2  ft.  of  slack 
in  R  which  must  be  drawn  over  to  F  in  addition  to  the  1  ft.  already  given  to  F  from  T. 
Therefore,  3  ft.  must  be  taken  up  at  F  for  every  foot  that  W  is  lifted.  Then  the 
mechanical  advantage  is  3. 


310 


ELEMENTS  OF  MECHANISM 


262.  Weston  Differential  Pulley  Block.  Fig.  391  shows  a  chain 
hoist  known  as  the  Weston  Differential  Pulley  Block.  The  two  upper 
sheaves  A  and  B  are  fast  to  each  other.  The  diameter  of  A  is  a  little 
larger  than  the  diameter  of  B  and  it  is  the  ratio  of  these  two  diameters 
which  governs  the  mechanical  advantage. 

The  diameter  of  the  lower  sheave  C  is  usually  a  mean  between  the 
diameters  of  the  upper  ones  in  order  that  the  supporting  chain  may 
hang  vertically.     This  feature  is  not  of  great  importance,  and  the 
diameter  of  the  lower  sheave  has  no  effect  on  the 
mechanical  advantage. 

The  chain  is  endless,  passing  over  A,  down  at  R, 
under  C,  up  at  P,  around  B,  and  hanging  loose. 
The  lifting  force  is  applied  at  F.  The  sheaves  are 
so  shaped  that  the  links  of  the  chain  fit  into  spaces 
provided  for  them  to  prevent  slipping. 

The  operation  of  the  hoist  may  be  seen  from  the 
following: 

Let  Da  represent  the  pitch  diameter  of  the 
sheave  A,  D6  the  pitch  diameter  of  the  sheave  B. 
Assume  that  the  chain  is  drawn  down  at  F  fast 
enough  to  cause  A  to  make  one  complete  turn  in 
a  unit  of  time;  that  is,  F  has  a  speed  of  TrDa  linear 
units.  This  would  give  an  upward  speed  to  the 
chain  at  R  of  irDa  linear  units.  Then,  if  B  were  not 
turning  the  sheave  C  would  roll  up  on  P,  its 
center  rising  at  a  speed  equal  to  one-half  the 
speed  of  the  chain  at  R',  that  is,  the  center  of  the  lower  sheave, 


FIG.  391 


and  therefore  the  weight  W,   would  rise  at  a  speed  of  —^  linear 

units.  But  at  the  same  time  that  R  is  rolling  C  up  on  P  the  pulley  B  is 
turning  at  the  same  angular  speed  as  A,  and  therefore  paying  out  chain 
at  P  at  the  rate  of  7rD&  linear  units  per  unit  of  time.  This  causes  C 
to  roll  down  on  R  at  a  speed  such  that  its  center  is  lowered  at  a  speed  of 

1^  linear  units.  The  resultant  upward  speed  of  the  center  of  C  is, 
therefore, 


7rD 


)&  =  7T  (Dg  -  D>) 

2 


2  2 

Since  the  speed  of  F  is  irDa  the  ratio  of  the  speed  of  F  to  that  of  W  is 

TrPg  2  Dg 

TT(Da-Db)   ~  Da-Db' 


MISCELLANEOUS  MECHANISMS 


311 


The  speed  ratio  may  be  found  grapnically  as  shown  in  Fig.  392. 
From  E  lay  off  along  the  chain  a  distance  V  representing  the  velocity 

of  E.    Draw  a  line  (shown  dotted)  from  the  end  of  this  distance,  to 

the  center  of  the  sheave.    The  length 

Vi  intercepted  on  the  line  of  the 

chain  through  E\  is  the  velocity  of 

Ei.     Draw  V\  downward  at  the  left- 
hand  side  of  the  lower  sheave  and  V 

upward  at  the  right-hand  side  of  the 

same  sheave.    Join  the  ends  of  these 

two  lines  as  shown,  getting  F4,  the 

resultant  velocity  of  M .    The  figure 

also  shows,  at  V%  and  V$,  the  effects 

of  V  and  V\  respectively,  when  as- 
sumed to  act  successively. 

263.   Intermittent  Motion  from 

Reciprocating  Motion.     A  recipro- 
cating  motion   in   one   piece   may 

cause  an   intermittent   circular  or 

rectilinear  motion  in  another  piece. 

It  may  be  arranged  that  one-half 

of  the  reciprocating  movement  is 

suppressed  and  that  the  other  half 

always  produces  motion  in  the  same 

direction,   giving  the  ratchet-wheel; 

or  the  reciprocating  piece  may  act  on  opposite  sides  of  a  toothed  wheel 

alternately,  and  allow  the  teeth  to  pass 
one  at  a  time  for  each  half  reciprocation, 
giving  the  different  forms  of  escapements 
as  applied  in  timepieces. 

264.  Ratchet-wheel.  A  wheel,  pro- 
vided with  suitably  shaped  pins  or  teeth, 
receiving  an  intermittent  circular  motion 
from  some  vibrating  or  reciprocating 
piece,  is  called  a  ratchet-wheel. 

In  Fig.  393,  A  represents  the  ratchet- 
wheel  turning  upon  the  shaft  a; '  C  is  an 

oscillating  lever  carrying  the  [detent,  click,  or  catch  B,  which  acts  on 

the  teeth  of  the  wheel.     The  whole  forms  the  three-bar  linkage  acb. 

When  the  arm   C   moves  left-handed,   the   click    B  will   push   the 

wheel  A   before  it  through  a  space  dependent   upon  the  motion  of 

C.    When  the  arm  moves  back,  the  click  will  slide  over  the  points 


FIG.  392 


FIG. 


312 


ELEMENTS  OF  MECHANISM 


of  the  teeth,  and  will  be  ready  to  push  the  wheel  on  its  forward 
motion  as  before;  in  any  case,  the  click  is  held  against  the  wheel 
either  by  its  weight  or  the  action  of  a  spring.  In  order  that  the 

arm  C  may  produce  motion  in  the  wheel 
A,  its  oscillation  must  be  at  least  suffi- 
cient to  cause  the  wheel  to  advance  one 
tooth. 

It  is  often  the  case  that  the  wheel  A 
must  be  prevented  from  moving  back- 
ward on  the  return  of  the  click  B.  In 
such  a  case  a  fixed  pawl,  click,  or  detent, 
similar  to  B,  turning  o  na  fixed  pin,  is  ar- 
ranged to  bear  on  the  wheel,  it  being 


FIG.  394. 


held  in  place  by  its  weight  or  a  spring.  Fig.  393  might  be  taken  to 
represent  a  retaining-pawl,  in  which  case  ac  is  a  fixed  link  and  the 
click  B  would  prevent  any  right  handed  motion  of  the  wheel  A. 
Fig.  394  shows  a  retaining-pawl  which  would  prevent  rotation  of  the 
wheel  A  in  either  direction;  such  pawls  are  often  used  to  retain  pieces 
in  definite  adjusted  positions. 

If  the  diameter  of  the  wheel  A  (Fig.  393)  be  increased  indefinitely, 
it  will  become  a  rack  which  would  then  receive  an  intermittent  trans- 
lation on  the  vibration  of  the  arm  C;  a  retaining-pawl  might  be  required 
in  this  case  also  to  prevent  a  backward  motion  of  the  rack. 

A  click  may  be  arranged  to  push,  as  in  Fig.  393,  or  to  pull,  as  in  Fig. 
400.  In  order  that  a  click  or  pawl  may  retain  its  hold  on  the  tooth  of  a 
ratchet-wheel,  the  common  normal 
to  the  acting  surfaces  of  the  click 
and  tooth,  or  pawl  and  tooth,  must 
pass  inside  of  .the  axis  of  a  pushing 
click  or  pawl,  as  shown  on  the 
lowest  click,  Fig.  395,  and  outside  the 
axis  of  the  pulling  click  or  pawl; 
the  normal  might  pass  through  the 
axis,  but  the  pawl  would  be  more 
securely  held  if  the  normal  is  located 
according  to  the  above  rule,  which 
also  secures  the  easy  falling  of  the 
pawl  over  the  points  of  the  teeth. 
It  is  sometimes  necessary,  or  more  convenient,  to  place  the  click- 
actuating  lever  on  an  axis  different  from  that  of  the  ratchet-wheel;  in 
such  a  case  care  must  be  taken  that  in  all  positions  of  the  click  the 
common  normal  occupies  the  proper  position;  it  will  generally  be  suffi- 


FIG.  395 


MISCELLANEOUS  MECHANISMS  313 

cient  to  consider  only  the  extreme  positions  of  the  pawl  in  any  case. 
Since  when  the  lever  vibrates  on  the  axis  of  the  wheel,  the  common 
normal  always  makes  the  same  angle  with  it  in  all  positions,  thus 
securing  a  good  bearing  of  the  pawl  on  the  tooth,  it  is  best  to  use  this 
construction  when  practicable. 

The  effective  stroke  of  a  click  or  pawl  is  the  space  through  which  the 
ratchet-wheel  is  driven  for  each  forward  stroke  of  the  arm.  The  total 
stroke  of  the  arm  should  exceed  the  effective  stroke  by  an  amount 
sufficient  to  allow  the  click  to  fall  freely  into  place. 

A  common  example  of  the  application  of  the  click  and  ratchet-wheel 
may  be  seen  in  several  forms  of  ratchet-drills  used  to  drill  metals  by 
hand.  As  examples  of  the  retaining-pawl  and  wheel  we  have  capstans 
and  windlasses,  where  it  is  applied  to  prevent  the  recoil  of  the  drum  or 
barrel,  for  which  purpose  it  is  also  applied  in  clocks. 

It  is  sometimes  desirable  to  hold  a  drum  at  shorter  intervals  than 
would  correspond  to  the  movement  of  one  tooth  of  the  ratchet-wheel; 
in  such  a  case  several  equal  pawls  may  be  used.  Fig.  395  shows  a 
case  where  three  pawls  were  used,  all  attached  by  pins  c,  Ci,  (%  to  the 
fixed  piece  C,  and  so  proportioned  that  they  come  into  action  alter- 
nately. Thus,  when  the  wheel  A  has  moved  an  amount  corresponding 
to  one-third  of  a  tooth,  the  pawl  B\  will  be  in  contact  with  the  tooth  61  ; 
after  the  next  one-third  movement,  BI  will  be  in  contact  with  62;  then 
after  the  remaining  one-third  movement,  B  will  come  into  contact  with 
the  tooth  under  6;  and  so  on.  This  arrangement  enables  us  to  obtain 
a  slight  motion  and  at  the  same  time  use  comparatively  large  and  strong 
teeth  on  the  wheel  in  place  of  small  weak  ones.  The  piece  C  might 
also  be  used  as  a  driving  arm,  and  the  wheel  could  then  be  moved 
through  a  space  less  than  that  of  a  tooth.  The  three  pawls  might  be 
made  of  different  lengths  and  placed  side  by  side  on  one  pin,  as  Ci,  in 
which  case  a  wide  wheel  would  be  necessary:  the  number  of  pawls 
required  would  be  fixed  by  the  conditions  in  each  case. 

265.  Reversible  Click  or  Pawl.  The  usual  form  of  the  teeth  of  a 
ratchet-wheel  is  that  given  in  Fig.  395,  which  only  admits  of  motion  in 
one  direction ;  but  in  feed  mechanisms,  such  as  those  in  use  on  shapers 
and  planers,  it  is  often  necessary  to  make  use  of  a  click  and  ratchet- 
wheel  that  will  drive  in  either  direction.  Such  an  arrangement  is  shown 
in  Fig.  396,  where  the  wheel  A  has  radial  teeth,  and  the  click,  which  is 
made  symmetrical,  can  occupy  either  of  the  positions  B  or  B',  thus  giv- 
ing to  A  a  right-  or  a  left-handed  motion.  In  order  that  the  click  B 
may  be  held  firmly  against  the  ratchet-wheel  A  in.  all  positions  of  the 
arm  C,  its  pivot  c,  after  passing  through  the  arm,  is  provided  with  a  small 
triangular  piece  (shown  dotted);  this  piece  turning  with  B  has  a  flat- 


314 


ELEMENTS  OF  MECHANISM 


ended  presser,  always  urged  upward  by  a  spring  (also  shown  dotted) 
bearing  against  the  lower  angle  opposite  B,  thus  urging  the  click  toward 
the  wheel;  a  similar. action  takes  place  when  the  click  is  in  the  dotted 
position  B'.  When  the  click  is  placed  in  line  with  the  arm  C,  it  is  held 
in  position  by  the  side  of  the  triangle  parallel  to  the  face  of  the  click; 
thus  this  simple  contrivance  serves  to  hold  the 
click  so  as  to  drive  in  either  direction,  and  also  to 
retain  it  in  position  when  thrown  out  of  gear. 

Since  for  different  classes  of  work  a  change  in  the 
"feed"  is  desired,  the  arrangement  must  be  such 
that  the  motion  of  the  ratchet-wheel  A  (Fig.  396), 
which  produces  the  feed,  can  be  adjusted.  This  is 
often  done  by  changing  the  swing  of  the  arm  C, 
which  is  usually  actuated  by  a  rod  attached  at  its 
free  end.  The  other  end  of  the  rod  is  attached  to  a 
vibrating  lever  which  has  a  definite  angular  move- 
ment at  the  proper  time  for  the  feed  to  occur,  and 
is  provided  with  a  T  slot  in  which  the  pivot  for  the 
rod  can  be  adjusted  by  means  of  a  thumb-screw  and  nut.  By  varying 
the  distance  of  the  nut  from  the  center  of  motion  of  the  lever,  the  swing 
of  the  arm  C  can  be  regulated;  to  reverse  the  feed,  it  occurring  in  the 
same  position  as  before,  the  click  must  be  reversed  and  the  nut  moved 
to  the  other  side  of  the  center  of  swing  of  the  lever. 

Figs.  397  and  398  show  other  methods  of  adjusting  the  motion  of  the 
ratchet-wheel.  In  Fig.  397,  which  shows  a  form  of  feed  mechanism 
used  by  Sir  J.  Whitworth  in  his  planing-machine,  C  is  an  arm  carrying 
the  click  B,  and  swinging  loosely  on  the  shaft  a  fixed  to  the  ratchet- 


FIG.  396 


FIG.  397 


FIG.  398 


wheel  A.  The  wheel  E,  also  turning  loosely  on  the  shaft  a,  and  placed 
just  behind  the  arm  C,  has  a  definite  angular  motion  sufficient  to  pro- 
duce the  coarsest  feed  desired;  its  concentric  slot  m  is  provided  with 
two  adjustable  pins  ee,  held  in  place  by  nuts  at  their  back  ends,  and 
enclosing  the  lever  C,  but  not  of  sufficient  length  to  reach  the  click  B. 
When  the  pins  are  placed  at  the  ends  of  the  slot,  no  motion  will  occur  in 


MISCELLANEOUS  MECHANISMS  315 

the  arm.  C;  but  when  e  and  e  are  placed  as  near  as  possible  to  each  other, 
confining  the  arm  C  between  them,  all  of  the  motion  of  E  will  be  given 
to  the  arm  C,  thus  producing  the  greatest  feed;  any  other  positions  of 
the  pins  will  give  motions  between  the  above  limits,  and  the  adjust- 
ment may  be  made  to  suit  each  case. 

In  Fig.  398,  the  stationary  shaft  a,  made  fast  to  the  frame  of  the 
machine  at  m,  carries  the  vibrating  arm  C,  ratchet-wheel  A,  and  ad- 
justable shield  S;  the  two  former  turn  loosely  on  the  shaft,  while  the 
latter  is  made  fast  to  it  by  means  of  a  nut  n,  the  hole  in  S  being  made 
smaller  than  that  in  A,  to  provide  a  shoulder  against  which  S  is  held 
by  the  nut.  The  arm  C  carries  a  pawl  B  of  a  thickness  equal  to  that  of 
the  wheel  plus  that  of  the  shield  S;  the  extreme  positions  of  this  pawl 
are  shown  by  dotted  lines  at  Bf  and  B".  The  teeth  of  the  wheel  A 
may  be  made  of  such  shape  as  to  gear  with  another  wheel  operating  the 
feed  mechanism;  or  another  wheel,  gearing  with  the  feed  mechanism, 
might  be  made  fast  to  the  back  of  A,  if  more  convenient:  in  the  latter 
case,  the  arm  C  would  be  placed  back  of  this  second  wheel. 

If  we  suppose  the  lever  in  its  extreme  left  position,  the  click  will  be 
at  B"  resting  upon  the  face  of  the  shield  S,  which  projects  beyond  the 
points  of  the  teeth  of  A ;  and  in  the  right-handed  motion  of  the  lever  the 
click  will  be  carried  by  the  shield  S  until  it  reaches  the  position  B,  where 
it  will  leave  the  shield  and  come  in  contact  with  the  tooth  6,  which  it  will 
push  to  &'  in  the  remainder  of  the  swing.  In  the  backward  swing  of 
the  lever  the  click  will  be  drawn  over  the  teeth  of  the  wheel  and  face  of 
the  shield  to  the  position  B".  In  the  position  of  the  shield  shown  in  the 
figure  a  feed  corresponding  to  three  teeth  of  the  wheel  A  is  produced; 
by  turning  the  shield  to  the  left  one,  two,  or  three  teeth,  a  feed  of  four, 
five,  or  six  teeth  might  be  obtained ;  while,  by  turning  it  to  the  right,  the 
feed  could  be  diminished,  the  shield  S  being  usually  made  large  enough  to 
consume  the  entire  swing  of  the  arm  C.  This  form  of  feed  mechanism 
has  often  been  used  in  slotting-machines,  and  in  such  cases,  as  well  as  in 
Figs.  397  and  398,  the  click  is  usually  held  to  its  work  by  gravity. 

266.  Double-acting  Click.  This  device  consists  of  two  clicks  mak- 
ing alternate  strokes,  so  as  to  produce  a  nearly  continuous  motion  of 
the  ratchet-wheel  which  they  drive,  that  motion  being  intermittent 
only  at  the  instant  of  reversal  of  the  movement  of  the  clicks.  In  Fig. 
399  the  clicks  act  by  pushing,  and  in  Fig.  400  by  pulling;  the  former 
arrangement  is  generally  best  adapted  to  cases  where  much  strength  is 
required,  as  in  windlasses. 

Each  single  stroke  of  the  click-arms  cdcf  (Fig.  399)  advances  the 
ratchet-wheel  through  one-half  of  its  pitch  or  some  multiple  of  its  half- 
pitch.  To  make  this  evident,  suppose  that  the  double  click  is  to  advance 


316 


ELEMENTS  OF  MECHANISM 


the  ratchet-wheel  one  tooth  for  each  double  stroke  of  the  click-arms, 
the  arms  being  shown  in  their  mid-stroke  position  in  the  figure.  Now 
when  the  click  be  is  beginning  its  forward  stroke,  the  click  b'c'  has  just 
completed  its  forward  stroke  and  is  beginning  its  backward  stroke; 
during  the  forward  stroke  of  be  the  ratchet-wheel  will  be  advanced  one- 
half  a  tooth;  the  click  b'c',  being  at  the  same  time  drawn  back  one- 
half  a  tooth,  will  fall  into  position  ready  to  drive  its  tooth  in  the  remain- 


FIG.  399 

ing  single  stroke  of  the  click-arms,  which  are  made  equal  in  length.  By 
the  same  reasoning  it  may  be  seen  that  the  wheel  can  be  moved  ahead 
some  whole  number  of  teeth  for  each  double  stroke  of  the  click-arms. 

In  Fig.  399  let  the  axis  a  and  dimensions  of  the  ratchet-wheel  be 
given,  also  its  pitch  circle  BB,  which  is  located  half-way  between  the 
tips  and  roots  of  the  teeth.  Draw  any  convenient  radius  ab,  and  from 
it  lay  off  the  angle  bae  equal  to  the  mean  obliquity  of  action  of  the  clicks, 

that  is,  the  angle  that  the  lines  of 
action  of  the  clicks  at  mid-stroke  are 
to  make  with  the  tangent  to  the  pitch 
circle  through  the  points  of  action. 
On  ae  let  fall  the  perpendicular  be, 
and  with  the  radius  ae  describe  the 
circle  CC:  this  is  the  base  circle,  to 
which  the  lines  of  action  of  the  clicks 
should  be  tangent.  Lay  off  the  angle 
eaf  equal,  to  an  odd  number  of  times 
the  half-pitch  angle,  and  through  the 
points  e  and  /,  on  the  base  circle, 
draw  two  tangents  cutting  each  other  in  h.  Draw  hd  bisecting  the  angle 
at  h,  and  choose  any  convenient  point  in  it,  as  d,  for  the  center  of  the 
rocking  shaft,  to  carry  the  click-arms.  From  d  let  fall  the  perpendiculars 
dc  and  dc'  on  the  tangents  hec  and  fhc'  respectively;  then  c  and  c'  will 
be  the  positions  of  the  click-pins,  and  dc  and  dc'  the  center  lines  of  the 


FIG.  400 


MISCELLANEOUS  MECHANISMS 


317 


click-arms  at  mid-stroke.  Let  b  and  b'  be  the  points  where  ce  and  c'f 
cut  the  pitch  circle;  then  cb  and  c'b'  will  be  the  lengths  of  the  clicks. 
The  effective  stroke  of  each  click  will  be  equal  to  half  the  pitch  as  meas- 
ured on  the  base  circle  CC  (or  some  whole  number  of  times  this  half- 
pitch),  and  the  total  stroke  must  be  enough  greater  to  make  the  clicks 
clear  the  teeth  and  drop  well  into  place. 

In  Fig.  400  the  clicks  pull  instead  of  push,  the  obliquity  of  action  is 
zero,  and  the  base  circle  and  pitch  circle  become  one,  the  points  6,  e, 
and  V,  f  (Fig.  399)  becoming  e  and  /  (Fig.  400).  In  all  other  respects 
the  construction  is  the  same  as  when  the  clicks  act  by  pushing,  and  the 
different  points  are  lettered  the  same  as  in  Fig.  399. 

Since  springs  are  liable  to  lose  their  elasticity  or  become  broken  after 
being  in  use  some  time,  it  is  often  desirable  to  get  along  without  apply- 
ing them  to  keep  clicks  in  position.  Fig.  401  shows  in  elevation  a 
mechanism  where  no  springs  are  required  to  keep  the  clicks  in  place,  it 


FIG.  401 


FIG.  402 


being  used  in  some  forms  of  lawn-mowers  to  connect  the  wheels  to  the 
revolving  cutter  when  the  mower  is  pushed  forward,  and  to  allow  a 
free  backward  motion  of  the  mower  while  the  cutter  still  revolves.  The 
ratchet  A  is  usually  made  on  the  inside  of  the  wheels  carrying  the  mower, 
and  the  piece  C,  turning  on  the  same  axis  as  A,  carries  the  three  equi- 
distant pawls  or  clicks  B,  shaped  to  move  in  the  cavities  provided  for 
them.  In  any  position  of  C,  at  least  one  of  the  clicks  will  be  held  in 
contact  with  A  by  the  action  of  gravity,  and  any  motion  of  A  in  the 
direction  of  the  arrow  will  be  given  to  the  piece  C.  Here  the  ratchet- 
wheel  drives  the  click,  ac  being  the  actuated  click-lever.  The  piece  C 
is  sometimes  attached  to  a  roller  by  means  of  the  shaft  a;  then  any 
left-handed  motion  of  C  will  be  given  to  A,  while  the  right-handed 
motion  will  simply  cause  the  clicks  to  slide  over  the  teeth  of  A.  The 
clicks  B  are  usually  held  in  place  by  a  cap  attached  to  C. 
Fig.  402  shows  a  form  of  click  which  is  always  thrown  into  action 


318 


ELEMENTS  OF  MECHANISM 


when  a  left-handed  rotation  is  given  to  its  arm  C,  while  any  motion  of 
the  wheel  A  left-handed  will  immediately  throw  the  click  out  of  action. 
The  wheel  A  carries  a  projecting  hub  d,  over  which  a  spring  D  is  fitted 
so  as  to  move  with  slight  friction.  One  end  of  this  spring  passes  between 
two  pins,  e,  placed  upon  an  arm  attached  to  the  click  B.  When  the 
arm  C  is  turned  left-handed,  the  wheel  A  and  the  spring  D  being  sta- 
tionary, the  click  B  will  be  thrown  toward  the  wheel  by  the  action  of 
the  spring  on  the  pin  e.  The  motion  of  the  wheel  A  will  be  equal  to 
that  of  the  arm  C,  minus  the  motion  of  C  necessary  to  throw  the  click 
into  gear.  Similarly,  when  A  turns  left-handed,  the  click  B  is  thrown 
out  of  gear.  This  mechanism  is  employed  in  some  forms  of  spinning- 
mules  to  actuate  the  spindles  when  winding  on  the  spun  yarn. 

267.  Friction-catch.  Various  forms  of  catches  depending  upon 
friction  are  often  used  in  place  of  clicks;  these  catches  usually  act  upon 
the  face  of  the  wheel  or  in  a  suitably  formed  groove  cut  in  the  face. 
Friction-catches  have  the  advantage  of  being  noiseless  and  allowing  any 
motion  of  the  wheel,  as  they  can  take  hold  at  any  point;  they  have  the 
disadvantage,  however,  of  slipping  when  worn,  and  of  getting  out  of 
order. 

Fig.  403  shows  a  friction-catch  B  working  in  a  V-shaped  groove  in 
the  wheel  A,  as  shown  in  section  A'B'.  Here  B  acts  as  a  retaining 

I 


FIG.  403 


FIG.  404 


click,  and  prevents  any  right-handed  motion  of  A ;  its  face  is  circular 
in  outline,  the  center  being  located  at  d,  a  little  above  the  axis  c.  A 
similarly  shaped  catch  might  be  used  in  place  of  an  actuating  click  to 
cause  motion  of  A. 

Fig.  404  shows  four  catches  like  B  (Fig.  403)  applied  to  drive  an 
annular  ring  A  in  the  direction  indicated  by  the  arrow.  When  the 
piece  c  is  turned  right-handed,  the  catches  B  are  thrown  against  the 
inside  b  of  the  annular  ring  by  means  of  the  four  springs  shown;  on 
stopping  the  motion  of  c,  the  pieces  B  are  pushed,  by  the  action  of  6, 


MISCELLANEOUS   MECHANISMS  319 

toward  the  springs  which  slightly  press  them  against  the  ring  and 
hold  them  in  readiness  to  again  grip  when  c  moves  right-handed.  Thus 
an  oscillation  of  the  piece  c  might  cause  continuous  rotation  of  the 
wheel  A,  provided  a  fly-wheel  were  applied  to  A  to  keep  it  going  while 
c  was  being  moved  back.  The  annular  ring  A  is  fast  to  a  disk  carried 
by  the  shaft  a;  the  piece  c  turning  loosely  on  a  has  a  collar  to  keep  it 
in  position  lengthwise  of  the  shaft. 

The  nipping-lever  shown  in  Fig.  405  is  another  application  of  the 
friction-catch.  A  loose  ring  C  surrounds  the  wheel  A ;  a  friction-catch 
B  having  a  hollow  face  works  in  a  pocket  in  the  ring  and  is  pivoted 
at  c.  On  applying  a  force  at  the  end  of  the  catch  B  in  the  direction 
of  the  arrow,  the  hollow  face  of  the  catch  will  "nip"  the  wheel  at  6, 
and  cause  the  ring  to  bear  tightly  against  the  left-hand  part  of  the  cir- 
cumference of  the  wheel;  the  friction  thus  set  up  will  cause  the  catch, 
ring,  and  wheel  to  move  together  as  one  piece.  The  greater  the  pull 
applied  at  the  end  of  the  catch  the  greater  will  be  the  friction,  as  the 
friction  is  proportional  to  the  pressure;  thus  the  amount  of  friction 
developed  will  depend  upon  the  resistance  to  motion  of  A.  Upon  re- 
versing the  force  at  the  end  of  the  catch,  the  hollow  face  of  the  catch 
will  be  drawn  away  from  the  face  of  A,  and  the  rounding  top  part 
of  the  catch,  coming  in  contact  with  the  top  of  the  cavity  in  the  ring, 
will  cause  the  ring  to  slide  back  upon  the  disk.  An  upward  motion  of 
the  click  end  will  again  cause  the  wheel  A  to  move  forward,  and  thus 
the  action  is  the  same  as  in  a  ratchet  and  wheel. 

Fig.  406  shows,  in  section,  a  device  which  has  been  applied  to  actu- 
ate sewing-machines  in  place  of  the  common  crank.  Two  such  mechan- 


FIG.  405  FIG.  406 

isms  were  used,  one  to  rotate  tne  shaft  of  the  machine  on  a  downward 
tip  of  the  treadle,  while  the  other  acted  during  the  upward  tip,  the 
treadle-rods  being  attached  to  the  projections  of  the  pieces  B.  The 
mechanism  shown  in  the  figure  acts  upon  the  shaft  during  the  down- 
ward motion  of  the  projection  B  as  shown  by  the  arrow. 

The  piece  C,  containing  an  annular  groove,  is  made  fast  to  the  shaft  a, 
the  sides  of  this  groove  being  turned  circular  and  concentric  with  the 


320 


ELEMENTS  OF  MECHANISM 


shaft.  The  piece  B,  having  a  projecting  hub  fitting  loosely  on  the 
inner  surface  of  the  groove  in  C,  is  placed  over  the  open  groove,  and 
is  held  in  place  by  a  collar  on  the  shaft.  The  hub  on  the  piece  B,  and 
the  piece  C,  are  shown  in  section.  The  friction-catch  D,  working  in  the 
groove,  is  fitted  over  the  hub  of  B,  the  hole  in  D  being  elongated  in 
the  direction  db  so  that  D  can  move  slightly  upon  the  hub  and  between 
the  two  pins  e  fast  in  the  piece  B.  A  cylindrical  roller  c  is  placed  in  the 
wedge-shaped  space  between  the  outer  side  of  the  groove  and  the  piece 
D,  a  spring  always  actuating  this  roller  in  a  direction  opposite  to  that 
of  the  arrow,  or  towards  the  narrower  part  of  the  space. 

Now  when  the  piece  B  is  turned  in  the  direction  of  the  arrow  by  a 
downward  stroke  of  the  treadle-rod,  it  will  move  the  piece  D  with  it  by 
means  of  the  pins  e;  at  the  same  time  the  roller  c  will  move  into  the 
narrow  part  of  the  wedge-shaped  space  between  C  and  D,  and  cause 
binding  between  the  pieces  D  and  C  at  b  and  at  the  surface  of  the  roller. 
The  friction  at  b  thus  set  up  will  cause  the  motion  of  D  to  be  given  to 
C.  On  the  upward  motion  of  the  projection  B  the  roller  will  be  moved 
to  the  large  part  of  its  space  by  the  action  of  the  piece  C  revolving  with 
the  shaft  combined  with  that  of  the  backward  movement  of  D,  thus 
releasing  the  pressure  at  b  and  allowing  C  to  move  freely  onward. 
The  other  catch  would  be  made  just  the  reverse  of  this  one,  and  would 
act  on  aniupward  movement  of  the  treadle-rod. 

Another  form  of  friction  catch,  some- 
times used  in  gang  saws  to  secure  the 
advance  of  the  timber  for  each  stroke  of 
the  saw,  and  called  the  silent  feed,  is  shown 
in  Fig.  407. 

The  saddle-block  B,  which  rests  upon 
the  outer  rim  of  the  annular  wheel  [A, 
carries  the  lever  C  turning  upon  the  pin  c. 
The  block  D,  which  fits  the  inner  rim  of 
the  wheel,  is  carried  by  the  lever  C,  and 
is  securely  held  to  its  lower  end  by  the  pin 
d  on  which  D  can  freely  turn.  When  the  pieces  occupy  the  positions 
shown  in  the  figure,  a  small  space  exists  between  the  piece  D  and  the 
inside  of  the  rim  A. 

The  upper  end  of  the  lever  C  has  a  reciprocating  motion  imparted  to 
it  by  means  of  the  rod  E.  The  oscillation  of  the  lever  about  the  pin  c 
is  limited  by  the  stops  e  and  G  carried  by  the  saddle-block  B.  When 
the  rod  E  is  moved  in  the  direction  indicated  by  the  arrow,  the  lever 
turning  on  c  will  cause  the  block  D  to  approach  B,  and  thus  nip  the  rim 
at  a  and  6;  and  any  further  motion  of  C  will  be  given  to  the  wheel  A. 


FIG.  407 


MISCELLANEOUS  MECHANISMS  321 

On  moving  E  in  the  opposite  direction  the  grip  will  first  be  loosened, 
and  the  lever  striking  against  the  stop  e  will  cause  the  combination  to 
slide  freely  back  on  the  rim  A .  The  amount  of  movement  given  to  the 
wheel  can  be  regulated  by  changing  the  stroke  of  the  rod  E  by  an 
arrangement  similar  to  that  described  in  connection  with  the  reversible 
click,  §  265.  The  stop  G  can  be  adjusted  by  means  of  the  screw  F  so 
as  to  prevent  the  oscillation  of  the  lever  upon  its  center  c,  thus  throw- 
ing the  grip  out  of  action.  The  saddle-block  B  then  merely  slides 
back  and  forth  on  the  rim,  the  action  being  the  same  as  that  obtained 
by  throwing  the  ordinary  click  out  of  gear. 

268.  Masked  Wheels.  It  is  sometimes  required  that  certain 
strokes  of  the  click-actuating  lever  shall  remain  inoperative  upon  the 
ratchet-wheel.  Such  arrangements  are  made  use  of  in  numbering- 
machines  where  it  is  desired  to  print  the  same  number  twice  in  succes- 
sion; they  are  called  masked  wheels. 

Fig.  408,  taken  from  a  model,  illustrates  the  action  of  a  masked 
wheel;  the  pin-wheel  D  represents  the  first  ratchet-wheel,  and  is  fast 
to  the  axis  a;  the  second  wheel  A  has  its  teeth 
arranged  in  pairs,  every  alternate  tooth  being 
cut  deeper,  and  it  turns  loosely  on  the  axis  a. 
The  click  B  is  so  made  that  one  of  its  acting 
surfaces,  i,  bears  against  the  pins  e  of  the  wheel 
D,  while  the  other,  g,  is  placed  so  as  to  clear  the 
pins  and  yet  bear  upon  the  teeth  of  A,  the  wheel 
A  being  located  so  as  to  admit  of  this. 

If  now  we  suppose  the  lever  C  to  vibrate 
through  an  angle  sufficient  to  move  either  wheel 
along  one  tooth,  both  having  the  same  number, 
it  will  be  noticed  that  when  the  projecting  piece  g 

is  resting  in  a  shallow  tooth  of  the  wheel  A,  the  acting  surface  i  will  be 
retained  too  far  from  the  axis  to  act  upon  the  tooth  e,  and  thus  this 
vibration  of  the  lever  will  have  no  effect  upon  the  pin-wheel  D;  while 
when  the  piece  g  rests  in  a  deep  tooth,  as  &',  the  click  will  be  allowed 
to  drop  so  as  to  bring  the  surface  i  into  action  with  the  pin  e'. 

In  the  figure  the  click  B  has  just  pushed  the  tooth  e'  into  its  present 
position,  the  projection  g  having  rested  in  the  deep  tooth  b'  of  the  wheel 
A ;  on  moving  back,  g  has  slipped  into  the  shallow  tooth  6,  and  thus  the 
next  stroke  of  the  lever  and  click  will  remain  inoperative  on  the  wheel 
D,  which  advances  but  one  tooth  for  every  two  complete  oscillations  of 
the  lever  C. 

Both  wheels  should  be  provided  with  retaining-pawls,  one  of  which, 
p,  is  shown.  This  form  of  pawl,  consisting  of  a  roller  p  turning  about 


322 


ELEMENTS  OF  MECHANISM 


a  pivot  carried  by  the  spring  s,  attached  to  the  frame  carrying  the  mech- 
anism, is  often  used  in  connection  with  pin-wheels,  as  by  rolling  between 
the  teeth  it  always  retains  them  in  the  same  position  relative  to  the  axis 
of  the  roller;  a  triangular-pointed  pawl  which  also  passes  between  the 
pins  is  sometimes  used  in  place  of  the  roller. 

The  pins  of  the  wheel  D  might  be  replaced  by  teeth  so  made  that 
their  points  would  be  just  inside  of  the  bottoms  of  the  shallow  teeth  of 
A;  a  wide  pawl  would  then  be  used,  and  when  it  rested  in  a  shallow 
tooth  of  A  it  would  remain  inoperative  on  D,  while  when  it  rested  in  a 
deep  tooth  it  would  come  in  contact  with  the  adjacent  tooth  of  D  and 
push  it  along. 

So  long  as  the  click  B  and  the  wheels  have  the  proper  relative  motion 
it  makes  no  difference  which  we  consider  as  fixed,  as  the  action  will  be 
the  same  whether  we  consider  the  axis  of  the  wheels  as  fixed  and  the 
click  to  move,  or  the  click  to  be  fixed  and  the  axis  to  have  the  proper 
relative  motion  in  regard  to  it.  The  latter  method  is  made  use  of  in 
some  forms  of  numbering-machines. 

269.  Counter  Mechanism.  Fig.  409  shows  the  mechanism  of  a 
"  counter "  used  to  record  the  number  of  double  strokes  made  by  a 
pump;  the  revolutions  made  by  a  steam  engine,  paddle,  propeller,  or 
other  shaft,  etc.  Two  views  are  given  in  the  figure,  which  represents 
a  counter  capable  of  recording  revolutions  from  1 
to  999;  if  it  is  desired  to  record  higher  numbers,  it 
will  only  be  necessary  to  add  more  wheels,  such  as 
A.  A  plate,  having  a  long  slot  or  series  of  openings 
opposite  the  figures  000,  is  placed  over  the  wheels, 
thus  only  allowing  the  numbers  to  be  visible  as  they 
come  under  the  slot  or  openings. 

The  number  wheels  A}  A\,  A%  are  arranged  to 
turn  loosely  side  by  side  upon  the  small  shaft  a, 
and  are  provided  with  a  series  of  ten  teeth  cut  into 
one  side  of  their  faces,  while  upon  the  other  side  a 
single  notch  is  cut  opposite  the  zero  tooth  on  the 
first  side,  it  having  the  same  depth  and  contour. 
This  single  notch  can  be  omitted  on  the  last  wheel  A%.  The  numbers 
0,  1,  2,  3,  4,  5,  6,  7,  8,  9  are  printed  upon  the  faces  of  the  wheels  in 
proper  relative  positions  to  the  teeth  t. 

Two  arms  C  are  arranged  to  vibrate  upon  the  shaft  a  of  the  number 
wheels,  and  carry  at  their  outer  ends  the  pin  c,  on  which  a  series  of 
clicks,  6,  61,  and  62,  are  arranged,  collars  placed  between  them  serving 
to  keep  them  in  position  on  the  pin.  The  arms  are  made  to  vibrate 
through  an  angle  sufficient  to  advance  the  wheels  one  tooth,  i.e.,  one- 


FIG.  409 


MISCELLANEOUS  MECHANISMS  323 

tenth  of  a  turn;  their  position  after  advancing  a  tooth  is  shown  by 
dotted  lines  in  the  side  view.  A  common  method  of  obtaining  this 
vibration  is  to  attach  a  rod  at  r,  one  end  of  the  pin  c,  this  rod  to  be  so 
attached  at  its  other  end  to  the  machine  as  to  cause  the  required  back- 
ward and  forward  vibrations  of  the  lever  C  for  each  double  stroke  or 
revolution  that  the  counter  is  to  record. 

The  click  6  is  narrow,  and  works  upon  the  toothed  edge  of  the  first 
wheel  A,  advancing  it  one  tooth  for  every  double  stroke  of  the  arm  c. 
The  remaining  clicks  61  and  62  are  made  broad,  and  work  on  the  toothed 
edges  of  Ai  and  Az,  as  well  as  on  the  notched  rims  of  A  and  A\,  re- 
spectively. When  the  notches  n  and  m  come  under  the  clicks  61  and  62 
the  clicks  will  be  allowed  to  fall  and  act  on  the  toothed  parts  of  AI  and 
A2;  but  in  any  other  positions  of  the  notches  the  clicks  will  remain 
inoperative  upon  the  wheels,  simply  riding  upon  the  smooth  rims  of  A 
and  Ai,  which  keep  the  clicks  out  of  action.  Each  wheel  is  provided 
with  a  retaining-spring  s  to  keep  it  in  proper  position. 

Having  placed  the  wheels  in  the  position  shown  in  the  figure,  the 
reading  being  000,  the  action  is  as  follows:  The  click  b  moves  the 
wheel  A  along  one  tooth  for  each  double  stroke  of  the  arm  C,  the  clicks 
bi  and  bz  remaining  inoperative  on  A\  and  A*;  on  the  figure  9  reaching 
the  slot,  or  the  position  now  occupied  by  0,  the  notch  n  will  allow  the 
click  61  to  fall  into  the  tooth  1  of  the  wheel  AI,  and  the  next  forward 
stroke  of  the  arm  will  advance  both  the  wheels  A  and  Ai,  giving  the 
reading  10;  the  notch  n  having  now  moved  along,  the  click  61  will 
remain  inoperative  until  the  reading  is  19,  when  61  will  again  come  into 
action  and  advance  AI  one  tooth,  giving  the  reading  20;  and  so  on  up 
to  90,  when  the  notch  m  comes  under  the  click  62.  To  prevent  the  click 
62  from  acting  on  the  next  forward  stroke  of  the  arm,  which  would 
make  the  reading  101  instead  of  91,  as  it  should  be,  a  small  strip  i  is 
fastened  firmly  to  the  end  of  the  click  62,  its  free  end  resting  upon  the 
click  61.  This  strip  prevents  the  click  62  from  acting  until  the  click  61 
falls,  which  occurs  when  the  reading  is  99;  on  the  next  forward  stroke 
the  clicks  61  and  62  act,  thus  giving  the  reading  100.  As  the  strip  merely 
rests  upon  61,  it  cannot  prevent  its  action  at  any  time.  If  another 
wheel  were  added,  its  click  would  require  a  strip  resting  on  the  end  of 
62.  A  substitute  for  these  strips  might  be  obtained  by  making  the 
wheel  A  fast  to  the  shaft  a,  and  allowing  the  remaining  wheels  to  turn 
loose  upon  it,  thinjdisks,  having  the  same  contour  as  the  notched  edge  of 
the  wheel  A,  being  placed  between  the  wheels  AiA2,  A2A3,  etc.,  and  made 
fast  to  the  shaft,  the  notches  all  being  placed  opposite  n;  thus  the  edges 
of  the  disks  would  keep  the  clicks  62,  63,  etc.,  out  of  action,  except  when 
the  figure  9  of  the  wheel  A  is  opposite  the  slot,  and  the  notches  m,  etc., 


324 


ELEMENTS  OF  MECHANISM 


are  in  proper  position.    A  simpler  form  of  counter  will  be  described  in 
§270. 

270.  Intermittent  Motion  from  Continuous  Motion.  The  cases  of 
intermittent  motion  thus  far  considered  have  been  those  in  which  a 
uniform  reciprocating  motion  in  one  piece  gives  an  intermittent  cir- 
cular or  rectilinear  motion  to  another,  the  click  being  the  driver  and 
the  wheel  the  follower. 

It  is  often  required  that  a  uniform  circular  motion  of  the  driver 
shall  produce  an  intermittent  circular  or  rectilinear  motion  of  the  fol- 
lower. The  following  examples  will  give  some  solutions  of  the  problem: 
Fig.  410  shows  a  combination  by  which  the  toothed  wheel  A  is 
moved  in  the  direction  of  the  arrow,  one  tooth  for  every  complete 
turn  of  the  shaft  d,  the  pawl  B  retaining  the  wheel  in  position  when  the 
tooth  t  on  the  shaft  d  is  out  of  action.  The  stationary  link  ado  forms 
the  frame,  and  provides  bearings  for  the  shafts  d  and  a,  and  a  pin  c  for 

the  pawl  B.  The  arm  e,  placed 
by  the  side  of  the  tooth  upon  the 
shaft,  is  arranged  to  clear  the 
wheel  A  in  its  motion,  and  to  lift 
the  pawl  B  at  the  time  when  the 
tooth  t  comes  into  action  wth  the 
wheel,  and  to  drop  the  pawl  when 
the  action  of  t  ceases,  i.e.,  when 
the  wheel  has  been  advanced  one 


FIG.  410 


tooth.  This  is  accomplished  by  attaching  the  piece  n  to  the  pawl,  its 
contour  in  the  raised  position  of  the  pawl  being  an  arc  of  a  circle  about 
the  center  of  the  shaft  d',  its  length  is  arranged  to  suit  the  above  re- 
quirements. When  the  tooth  t  comes  in  contact  with  the  wheel,  the 
arm  e,  striking  the  piece  n,  raises  the  pawl  (which  is  held  in  position  by 
the  spring  s),  and  retains  it  in  the  raised  position  until  the  tooth  t  is 
ready  to  leave  the  wheel,  when  e,  passing  off  from  the 
end  of  n,  allows  the  pawl  to  drop. 

In  Fig.  411  the  wheel  A  makes  one-third  of  a  revolu- 
tion for  every  turn  of  the  wheel  be,  its  period  of  rest 
being  about  one-half  the  period  of  revolution  of  be. 
If  we  suppose  A  the  follower,  and  to  turn  right-handed 
while  the  driver  be  turns  left-handed,  one  of  the  round 
pins  b  is  just  about  to  push  ahead  the  long  tooth  of  A, 
the  circular  retaining  sector  c  being  in  such  a  position  as  to  follow  a 
right-handed  motion  of  A .  The  first  pin  slides  down  the  long  tooth,  and 
the  other  pins  pass  into  and  gear  with  the  teeth  b',  the  last  pin  passing 
off  on  the  long  tooth  e,  when  the  sector  c  will  come  in  contact  with  the 


FIG.  411 


MISCELLANEOUS  MECHANISMS 


325 


arc  c',  and  retain  the  wheel  A  until  the  wheel  be  again  reaches  its 
present  position. 

Fig.  412  is  a  diagram  of  a  mechanism  known  as  a  Geneva  stop. 
The  wheel  A  makes  one-sixth  of  a  revolution  for  one  turn  of  the  driver 
ac,  the  pin  6  working  in  the  slots  &'  causing  the  motion  of  A ;  while  the 
circular  portion  c  of  the  driver,  coming  in  contact  with 
the  corresponding  circular  hollows  c',  retains  A  in  position 
when  the  tooth  b  is  out  of  action.  The  wheel  a  is  cut 
away  just  back  of  the  pin  b  to  provide  clearance  for  the 
wheel  A  in  its  motion.  If  we  close  up  one  of  the  slots, 
as  &',  it  will  be  found  that  the  shaft  a  can  only  make  a 
little  over  five  and  one-half  revolutions  in  either  direction 
before  the  pin  6  will  strike  the  closed  slot.  This 
mechanism,  when  so  modified,  has  been  applied  to 
watches  to  prevent  overwinding,  and  is  called  the  Geneva  stop,  the 
wheel  a  being  attached  to  the  spring-shaft  so  as  to  turn  with  it,  while 
A  turns  on  an  axis  d  in  the  spring-barrel.  The  number  of  slots  in  A 
depends  upon  the  number  of  times  it  is  desired  to  turn  the  spring- 
shaft. 

By  placing  another  pin  opposite  b  in  the  wheel  ac,  as  shown  by  dotted 
lines,  and  providing  the  necessary  clearance,  the  wheel  A  could  be 
moved  through  one-sixth  of  a  turn  for  every  half  turn  of  ac. 

A  simple  type  of  counter  extensively  used  on  water-meters  is  shown 
in  Fig.  413.  It  consists  of  a  series  of  wheels  A,  B,  C,  mounted  side  by 


FIG.  412 


FIG.  413 


side  and  turning  loosely  on  the  shaft  /S>;  or  the  first  wheel  to  the  right 
may  be  fast  to  the  shaft  and  all  the  remaining  wheels  loose  upon  it. 
Each  wheel  is  numbered  on  its  face  as  in  Fig.  409,  and  it  is  provided,  as 
shown,  that  the  middle  row  of  figures  appears  in  a  suitable  slot  in  the 
face  of  the  counter.  The  first  wheel  A  is  attached  to  the  worm-wheel  E, 


326 


ELEMENTS  OF  MECHANISM 


having  20  teeth  and  driven  by  the  worm  F  geared  to  turn  twice  for  one 
turn  of  the  counter  driving  shaft. 

On  a  parallel  shaft  T  loose  pinions  D  are  arranged  between  each 
pair  of  wheels.  Each  pinion  is  supplied  with  six  teeth  on  its  left  side 
extending  over  a  little  more  than  one-half  its  face  and  with  three  teeth, 
each  alternate  tooth  being  cut  away,  for  the  remainder  of  the  face,  as 
clearly  shown  in  the  sectional  elevations.  The  middle  elevation  (Fig. 
413)  shows  a  view  of  the  wheel  B  from  the  right  of  the  line  ab  with  the 
pinion  D  sectioned  on  the  line  cd.  The  right  elevation  shows  a  view  of 
the  wheel  A  from  the  left  of  the  line  ab  with  the  pinion  D  sectioned  on 
the  line  cd.  The  first  wheel  A,  and  all  others  except  the  last,  at  the 
left,  have  on  their  left  sides  a  double  tooth  G,  which  is  arranged  to  come 
in  contact  with  the  six-tooth  portion  of  the  pinion;  the  space  between 
these  teeth  is  extended  through  the  brass  plate  which  forms  the  left 
side  of  the  number  ring  whose  fperiphery  H  acts  as  a  stop  for  the  three- 
tooth  portion  of  the  pinion,  as  clearly  shown  in  the  figure  to  the  right. 
Similarly  on  the  right  side  of  each  wheel,  except  the  first,  is  placed  a 
wheel  of  20  teeth  gearing  with  the  six-tooth  part  of  the  pinion,  as  shown 
in  the  middle  figure.  When  the  digit  9  on  any  wheel,  except  the  one  at 
the  left,  comes  under  the  slot,  the  double  tooth  G  is  ready  to  come  in 
contact  with  the  pinion;  as  the  digit  9  passes  under  the  slot  the  tooth 
G  starts  the  pinion,  which  is  then  free  to  make  one-third  of  a  turn  and 
again  become  locked  by  the  periphery  H.  Thus  any  wheel  to  the  left 
receives  one-tenth  of  a  turn  for  every  passage  of  the  digit  9  on  the  wheel 
to  its  right."  In  the  figure  the  reading  329  will  change  to  330  on  the 
passing  of  the  digit  9.  This  counter  can  be  made  to  record  oscillations 
by  supplying  its  actuating  shaft  with  a  ten-tooth  ratchet,  arranged  with 
a  click  to  move  one  tooth  for  each  double  oscillation. 

Figs.  414a  and  414b  show  two  methods  of  advancing  the  wheels  A 
through  a  space  corresponding  to  one  tooth  during  a  small  part  of  a 
revolution  of  the  shafts  c;  in  this  case  the  shafts  are  at  right  angles  to 
each  other.  In  Fig.  414a  a  raised  circular  ring  with  a  small  spiral 
ft. 


FIG.  414a 


FIG.  414b 


part  b  attached  to  a  disk  is  made  use  of;  the  circular  part  of  the  ring 
retains  the  wheel  in  position,  while  the  spiral  part  gives  it  its  motion. 
In  Fig.  414b  the  disk  carried  by  the  shaft  cc  has  a  part  of  its  edge  bent 


MISCELLANEOUS  MECHANISMS 


327 


helically  at  6;  this  helical  part  gives  motion  to  the  wheel,  and  the 
remaining  part  of  the  disk  edge  retains  the  wheel  in  position.  By 
using  a  regular  spiral,  in  Fig.  414a,  and  one  turn  of  a  helix,  in  Fig. 
414b,  the  wheels  A  could  be  made  to  move  uniformly  through  the 
space  of  one  tooth  during  a  uniform  revolution  of  the  shafts  c. 

In  Fig.  415  the  wheel  A  is  arranged  to  turn  the  wheel  B,  on  a  shaft 
at  right  angles  to  that  of  A,  through  one-half  a  turn  while  it  turns  one- 
sixth  of  a  turn,  and  to 
lock  B  during  the  remain- 
ing five-sixths  of  the  turn. 
Fig.  416  illustrates  the 
Star  Wheel.     The  wheel 

)A,  turns  through  a  space 
corresponding  to  one 
tooth  for  each  revolution 
of  the  arm  carrying  the 
pin  b  and  turn- 
ing on  the  shaft 
c.  The  pin  b  is 
often  stationary, 
and  the  star 
wheel  is  moved 
past  it;  theFiQ  416 
action  is  then 

evidently  the  same,  as  the  pin  and  wheel  have  the  same  relative  motion 
in  regard  to  each  other  during  the  time  of  action.  The  star  wheel  is 
often  used  on  moving  parts  of  machines  to  actuate  some  feed  mechan- 
ism, as  may  be  seen  in  cylinder- 
boring  machines  on  the  facing 
attachment,  and  in  spinning- 
machinery. 

271.  Cam  and  Slotted 
Sliding  Bar.  Fig.  417  shows 
an  equilateral  triangle  abc, 
formed  by  three  circular  arcs, 
whose  centers  are  at  a,  6,  and  c, 
the  whole  turning  about  the 
axis  a,  and  producing  an  inter- 
mittent motion  in  the  slotted 
piece  B.  The  width  of  the  slot  is  equal  to  the  radius  of  the  three  circular 
arcs  composing  the  three  equal  sides  of  the  triangular  cam  A,  and  there- 
fore the  cam  will  always  bear  against  both  sides  of  the  groove. 


FIG.  415 


FIG.  417 


FIG.  418 


328  ELEMENTS  OF  MECHANISM 

If  we  imagine  the  cam  to  start  from  the  position  shown  in  Fig.  418 
when  6  is  at  1,  the  slotted  piece  B  will  remain  at  rest  while  b  moves 
from  1  to  2  (one-sixth  of  the  circle  1,  2  .  .  .  6),  the  cam  edge  be  merely 
sliding  over  the  lower  side  of  the  slot.  When  6  moves  from  2  to  3,  i.e., 
from  the  position  of  A,  shown  by  light  full  lines,  to  that  shown  by  dotted 
lines,  the  edge  ab  will  act  upon  the  upper  side  of  the  slot,  and  impart 
to  B  a  motion  similar  to  that  obtained  in  Fig.  357,  being  that  of  a 
crank  with  an  infinite  connecting-rod;  from  3  to  4  the  point  b  will  drive 
the  upper  side  of  the  slot,  ca  sliding  over  the  lower  side,  the  motion  here 
being  also  that  of  a  connecting-rod  with  an  infinite  link,  but  decreasing 
instead  of  increasing  as  from  2  to  3.  When  b  moves  from  4  to  5  there 
is  no  motion  in  B;  from  5  to  6,  c  acts  upon  the  upper  side  of  the  slot, 
and  B  moves  downward;  from  6  to  1,  ac  acts  on  the  upper  side  of  the  slot, 
and  B  moves  downward  to  its  starting  position.  The  motion  of  B  is 
accelerated  from  5  to  6  and  retarded  from  6  to  1. 

At  A'  a  form  of  cam  is  shown  where  the  shaft  a  is  wholly  contained 
in  the  cam.  In  this  case  draw  the  arcs  de  and  cb  from  the  axis  of  the 
shaft  as  a  center,  making  ce  equal  to  the  width  of  the  slot  in  B;  from  c 
as  a  center  with  a  radius  ce  draw  the  arc  eb,  and  note  the  point  b  where 
it  cuts  the  arc  c6;  with  the  same  radius  and  b  as  a  center  draw  the  arc 
dc,  which  will  complete  the  cam.  In  this  case  the  angle  cab  will  not  be 
equal  to  60°,  and  the  motions  in  their  durations  and  extent  will  vary  a 
little  from  those  described  above. 

272.  Locking  Devices.  The  principle  of  the  slotted  sliding  bar 
combined  with  that  of  the  Geneva  stop  is  applied  in  the  shipper  mechan- 
ism shown  in  Fig.  419,  often  used  on  machines  where  the  motion  is 

automatically  reversed.  The  shipper  bar  B 
slides  in  the  piece  CC,  which  also  provides  a 
pivot  a  for  the  weighted  lever  wab.  The  end 
of  the  lever  b  opposite  the  weight  w  carries  a 
plG  41g  pin  which  works  in  the  grooved  lug  s  on  the 

shipper  bar.     In  the  present  position  of  the 

pieces,  the  pin  b  is  in  the  upper  part  of  the  slot,  and  the  weight  w,  tending 
to  fall  under  the  action  of  gravity,Jholds  it  there,  the  shipper  being  thus 
effectually  locked  in  its  present  position.  If  now  the  lever  be  turned  left- 
handed  about  its"axis  a  until  the  weight  w  is  just  a  little  to  the  left  of  a, 
gravity  will  carry  the  weight  and  lever  into  the  dotted  position  shown, 
where  it  will  be  locked  until  the  lever  is  turned  right-handed.  The  princi- 
ple of  using  a  weight  to  complete  the  motion  is  very  convenient,  as  the  part 
of  the  machine  actuating  the  shipper  often  stops  before  the  belt  is 
carried  to  the  wheel  which  produces  the  reverse  motion,  and  the  machine 
is  thus  stopped.  The  motion  can  always  be  made  sufficient  to  raise  a 


MISCELLANEOUS  MECHANISMS  329 

weighted  lever,  as  shown  above,  and  the  weight  will,  in  falling,  com- 
plete the  motion  of  the  shipper. 

The  device  shown  in  Fig.  420,  of  which  there  may  be  many  forms, 
serves  to  retain  a  wheel  A  in  definite  adjusted  positions,  its  use  being 
the  same  as  that  of  the  retaining-pawl  shown  in  Fig.  394.  The  wheels 
B  and  A  turn  on  the  shafts  c  and  a,  respectively,  carried  by  the  link  C, 
which  is  shown  dotted,  as  it  has  been  cut  away  in  taking  the  section. 
Two  positions  of  the  wheel  B  will  allow  the  teeth  b  of  A  to  pass  freely 
through  its  slotted  opening,  while  any  other  position  effectually  locks 
the  wheel  A.  The  shape  of  the  slot  in  B  and  the  teeth  of  A  are  clearly 
shown  in  the  figure. 

Fig.  421  shows  another  device  for  locking  the  wheel  A,  the  teeth  of 
which  are  round  pins;  but  in  this  case  it  is  necessary  to  turn  B  once  to 
pass  a  tooth  of  A.  If  we  suppose  the  wheel  A  under  the  influence  of 


FIG.  420  FIG.  421 

a  spring  which  tends  to  turn  it  right-handed,  and  then  turn  B  uniformly 
either  right-  or  left-handed,  the  wheel  A  will  advance  one  tooth  for  each 
complete  turn  of  B}  a  pin  first  slipping  into  the  groove  on  the  left  and 
leaving  it  when  the  groove  opens  toward  the  right,  the  next  pin  then 
coming  against  the  circular  part  of  B  opposite  the  groove.  It  will  be 
noticed  that  while  there  are  only  six  pins  on  the  wheel  A,  yet  there  are 
twelve  positions  in  which  A  can  be  locked,  as  a  tooth  may  be  in  the 
bottom  of  the  groove  or  two  teeth  may  be  bearing  against  the  circular 
outside  of  B.  Devices  similar  in  principle  to  those  shown  in  Figs.  420 
and  421  are  often  used  to  adjust  stops  in  connection  with  feed 
mechanisms. 

Clicks  and  pawls  as  used  in  practice  may  have  many  different  forms 
and  arrangements;  their  shape  depends  very  much  upon  their  strength 
and  the  space  in  which  they  are  to  be  placed,  and  the  arrangement 
depends  on  the  requirements  in  each  case. 


330  ELEMENTS  OF  MECHANISM 

273.  Escapements.     An  escapement  is  a  combination  in  which  a 
toothed  wheel  acts  upon  two  distinct  pieces  or  pallets  attached  to  a  re- 
ciprocating frame,  it  being  so  arranged  that  when  one  tooth  escapes  or 
ceases  to  drive  its  pallet,  another  tooth  shall  begin  its  action  on  the 
other  pallet. 

A  simple  form  of  escapement  is  shown  in  Fig.  422.     The  frame  cc' 
is  arranged  to  slide  longitudinally  in  the  bearings  CC,  which  are  at- 
tached to  the  bearing  for  the  toothed 
wheel.    The  wheel  a  turns  continually 
in  the  direction  of  the  arrow,  and  is 
provided  with  three  teeth,  6,  &',  6",  the 
frame  having  two  pallets,  c  and  c'.    In 
the  position  shown,  the  tooth  b  is  just 
422  ceasing  to  drive  the  pallet  c  to  the  right, 

and  is  escaping,  while  the  tooth  b'  is 

just  coming  in  contact  with  the  pallet  c',  when  it  will  drive  the  frame 
to  the  left. 

While  escapements  are  generally  used  to  convert  circular  into  recip- 
rocating motion,  as  in  the  above  example,  the  wheel  being  the  driver, 
yet,  in  many  cases,  the  action  may  be  reversed.  In  Fig.  422,  if  we 
consider  the  frame  to  have  a  reciprocating  motion  and  use  it  as  the 
driver,  the  wheel  will  be  made  to  turn  in  the  opposite  direction  to  that 
in  which  it  would  itself  turn  to  produce  reciprocating  motion  in  the 
frame.  It  will  be  noticed  also  that  there  is  a  short  interval  at  the 
beginning  of  each  stroke  of  the  frame  in  which  no  motion  will  be  given 
to  the  wheel.  It  is  clear  that  the  wheel  a  must  have  1,  3,  5,  or  some 
odd  number  of  teeth  upon  its  circumference. 

274.  The   Crown-wheel   Escapement.    The   crown-wheel   escape- 
ment (Fig.  423)  is  used  for  causing  a  vibration  in  one  .axis  by  means  of 
a  rotation  of  another.     The  latter  carries  a  crown 

wheel  A,  consisting  of  a  circular  band  with  an  odd 

number  of  large  teeth,  like  those  of  a  splitting-saw, 

cut  on  its  upper  edge.     The  vibrating  axis,  o,  or 

verge  as  it  is  often  called,  is  located  just  above  the 

teeth  of  the  crown  wheel,  in  a  plane  at  right  angles 

to  the  vertical  wheel  axis.     The  verge  carries  two 

pallets,  b  and  61,  located  in  planes  passing  through 

its  axis,  the  distance  between  them  being  arranged 

so  that  they  may  engage  alternately  with  teeth  on  opposite  sides  of 

the  wheel.     If  the  crown  wheel  be  made  to  revolve  under  the  action 

of  a  spring  or  weight,  the  alternate  action  of  the  teeth  on  the  pallets 

will  cause  a  reciprocating  motion  in  the  verge.    The  rapidity  of  this 


MISCELLANEOUS  MECHANISMS 


331 


vibration  depends  upon  the  inertia  of  the  verge,  which  may  be  adjusted 
by  attaching  to  it  a  suitably  weighted  arm. 

This  escapement,  having  the  disadvantage  of  causing  a  recoil  in  the 
wheel  as  the  vibrating  arm  cannot  be  suddenly  stopped,  is  not  used  in 
timepieces,  and  but  rarely  in  other  places.  It  is  of  interest,  however, 
as  being  the  first  contrivance  used  in  a  clock  for  measuring  time. 

275.  The  Anchor  Escapement.  The  anchor  escapement  as  applied 
in  clocks  is  shown  in  Fig.  424.  The  escape-wheel  AI  turns  in  the  direc- 
tion of  the  arrow  and  is  supplied  with  long  pointed  teeth.  The  pallets 
are  connected  to  the  vibrating  axis  or  verge  Ci  by  means  of  the  arms 
and  eiCi,  the  axis  of  the  verge  and  wheel  being  parallel  to  each 


FIG.  424 

other.  The  verge  is  supplied  at  its  back  end  with  an  arm  C\pi,  carry- 
ing a  pin  pi  at  its  lower  end.  This  pin  works  in  a  slot  in  the  pendulum- 
rod,  not  shown.  The  resemblance  of  the  two  pallet  arms  combined 
with  the  upright  arm  to  an  anchor  gave  rise  to  the  name  "anchor 
escapement."  The  left-hand  pallet,  d\,  is  so  shaped  that  all  the  nor- 
mals to  its  surface  pass  above  the  verge  axis  Ci,  while  all  the  normals 
to  the  right-hand  pallet,  e\,  pass  below  the  axis  C\.  Thus  an  upward 
movement  of  either  pallet  will  allow  the  wheel  to  turn  in  the  direction 
of  the  arrow,  or,  the  wheel  turning  in  the  direction  of  the  arrow,  will, 
when  the  tooth  61  is  in  contact  with  the  pallet  di,  cause  a  left-handed 
swing  of  the  anchor;  and  when  61  has  passed  off  from  di  and  01  reaches 
the  right-hand  pallet,  as  shown,  a  right-handed  swing  will  be  given  to 
the  anchor.  As  the  pendulum  cannot  be  suddenly  stopped  after  a 


332  ELEMENTS  OF  MECHANISM 

tooth  has  escaped  from  a  pallet,  the  tooth  that  strikes  the  other  pallet 
is  subject  to  a  slight  recoil  before  it  can  move  in  the  proper  direction, 
which  motion  begins  when  the  pendulum  commences  its  return  swing. 
The  action  of  the  escape-wheel  on  the  pendulum  is  as  follows : 

Suppose  the  points  l\  and  k\  to  show  extreme  positions  of  the  point  p\, 
and  suppose  the  pendulum  and  point  p\  to  be  moving  to  the  left;  the 
tooth  61  has  just  escaped  from  the  pallet  d\,  and  01  has  impinged  on  e\, 
as  shown,  the  point  p\  having  reached  the  position  m\.  The  recoil 
now  begins,  the  pallet  e{  moving  back  the  tooth  01,  while  pt  goes  from 
mi  to  l\.  The  pendulum  then  swings  to  the  right  and  the  pallet  e\  is 
urged  upward  by  the  tooth  01,  thus  urging  the  pendulum  to  the  right 
while  pi  passes  from  l\  to  n\t  when  Oi  escapes.  Recoil  then  occurs  on 
the  pallet  di  from  n\  to  ki,  and  from  k\  to  mi  an  impulse  is  given  to  the 
pendulum  to  the  left,  when  the  above-described  cycle  will  be  repeated. 
As  the  space  through  which  the  pendulum  is  urged  on  exceeds  that 
through  which  it  is  held  back,  the  action  of  the  escape-wheel  keeps  the 
pendulum  vibrating.  This  alternate  action  with  and  against  the  pen- 
dulum prevents  it  from  being,  as  it  should  be,  the  sole  regulator  of  the 
speed  of  revolution  of  the  escape- wheel;  for  its  own  time  of  vibratkm, 
instead  of  depending  only  upon  its  length,  will  also  depend  upon  the 
force  urging  the  escape-wheel  round.  Therefore  any  change  in  the 
maintaining  force  will  disturb  the  rate  of  the  clock. 

276.  Dead-beat  Escapement.  The  objectionable  feature  of  the 
anchor  escapement  is  removed  in  Graham's  dead-beat  escapement, 
shown  in  Fig.  425.  The  improvement  consists  in  making  the  outline  of 
the  lower  surface,  db,  of  the  left-hand  pallet,  and  the  upper  surface  of 
the  right-hand  pallet,  arcs  of  a  circle  about  (7,  the  verge  axis;  the 
oblique  surfaces  b  and  /  complete  the  pallets.  The  construction  indi- 
cated by  dotted  lines  in  the  figure  insures  that  the  oblique  surfaces  of 
the  pallets  shall  make  equal  angles,  in  their  normal  position,  with  the 
tangents  bC  and  fC  to  the  wheel  circle  not  shown.  If  we  suppose  the 
limits  of  the  swing  of  the  point  p  to  be  I  and  k,  the  action  of  the  escape- 
wheel  on  the  pendulum  is  as  follows: 

The  pendulum  being  in  its  right  extreme  position,  the  tooth  b  is 
bearing  against  the  circular  portion  of  the  pallet  d;  as  the  pendulum 
swings  tp  the  left  under  the  action  of  gravity,  the  tooth  b  will  begin  to 
move  along  the  inclined  face  of  the  pallet  when  the  center  line  has 
reached  n,  and  will  urge  the  pendulum  onward  to  m,  where  the  tooth 
leaves  the  pallet,  and  another  tooth  o  comes  in  contact  with  the  cir- 
cular part  of  the  pallet  e,  which,  with  the  exception  of  a  slight  friction 
between  it  and  the  point  of  the  tooth,  will  leave  the  pendulum  free  to 
move  onward,  the  wheel  being  locked  in  position.  On  the  return 


MISCELLANEOUS  MECHANISMS 


333 


swing  of  the  pendulum,  the  inclined  part  of  the  pallet  e  urges  the  pen- 
dulum from  m  to  n.  Hence  there  is  no  recoil,  and  the  only  action 
against  the  pendulum  is  the  very  minute  friction  between  the  teeth 
and  the  pallets.  The  impulse  is  here  given  through  an  arc  mn,  very 
nearly  bisected  by  the  middle  point  of  the  swing  of  the  pendulum, 
which  is  also  an  advantage.  The  term  " dead-beat"  has  been  applied 


FIG.  425 

because  the  second  hand,  which  is  fitted  to  the  escape-wheel,  stops  so 
completely  when  the  tooth  falls  upon  the  circular  portion  of  a  pallet, 
there  being  no  recoil  or  subsequent  trembling  such  as  occurs  in  other 
escapements. 

In  watches  the  pendulum  is  replaced  by  a  balance-wheel  swinging 
backward  and  forward  on  an  arbor  under  the  action  of  a  very  light 
coiled  spring,  often  called  a  "  hair-spring "  the  pivots  of  the  arbor  are 
very  nicely  made,  so  that  they  turn  with  very  slight  friction. 

277.  The  Graham  Cylinder  Escapement.  This  form 
of  escapement  is  used  in  the  Geneva  watches.  Here  the 
balance  verge  o  (Fig.  426)  has  attached  to  it  a  very  thin 
cyclindrical  shell  rs  centered  at  o,  the  axis  of  the  verge,  and 
the  point  of  the  tooth  b  can  rest  either  on  the  outside  or 
inside  of  the  cylinder  during  a  part  of  the  swing  of  the  bal- 
ance. As  the  cylinder  turns  in  the  direction  of  the  arrow 
(Fig.  426a),  the  wheel  also  being  urged  in  the  direction  of 
its  arrow,  the  inclined  surface  of  the  tooth  be  comes  under 
the  edge  s  of  the  cylinder,  and  thus  urges  the  balance  onward ;  this  gives 
one  impulse,  as  shown  in  Fig.  426b.  The  tooth  then  passes  s,  flies  into 


FIG.  426a 


FIG.  426b 


334 


ELEMENTS  OF   MECHANISM 


the  cylinder,  and  is  stopped  by  the  concave  surface  near  r.  In  the 
opposite  swing  of  the  balance  the  tooth  escapes  from  the  cylinder,  the 
inclined  surface  pushing  r  upward,  which  gives  the  other  impulse  in  the 
opposite  direction  to  the  first;  the  action  is  then  repeated  by  the  next 
tooth  of  the  wheel. 

This  escapement  is,  in  its  action,  nearly  identical  to  the  dead-beat; 
but  the  impulse  is  here  given  through  small  equal  arcs,  situated  at 
equal  distances  from  the  middle  point  of  the  swing. 

278.  The  Chronometer  Escapement  is  shown  in  Fig.  427.  Here 
the  verge  o  carries  two  circular  plates,  one  of  which  carries  a  projec- 
tion p,  which  serves  to  operate  the  detent  d;  the  other  carries  a  pro- 
jection n,  which  swings  freely  by  the  teeth  of  the  escape-wheel  when  a 


FIG.  427 

tooth  is  resting  upon  the  pallet  d,  but  encounters  a  tooth  when  the 
wheel  is  in  any  other  position. 

The  detent  d  has  a  compound  construction  and  consists  of  four 
parts : 

1°  The  locking-stone  d,  a  piece  of  ruby  on  which  the  tooth  of  the 
escape-wheel  rests. 

2°  The  discharging-spring  Z,  a  very  fine  strip  of  hammered  gold. 

3°  A  spring  s  on  which  the  detent  swings,  and  which  attaches  the 
whole  to  the  frame  of  the  chronometer. 

4°  A  support  e,  attached  to  the  body  of  the  detent,  to  prevent  the 
strip  I  from  bending  upward. 

A  pin  r  prevents  the  detent  from  approaching  too  near  the  wheel. 

The  action  of  the  escapement  is  as  follows:  On  a  right-hand  swing 
of  the  balance  the  projection  p  meets  the  light  strip  I,  which,  bending 
from  its  point  of  attachment  to  the  detent,  offers  but  very  little  resist- 


MISCELLANEOUS  MECHANISMS  335 

ance  to  the  balance.  On  the  return  swing  of  the  balance,  the  projec- 
tion p  meets  the  strip  I,  which  can  now  only  bend  from  e,  and  raises 
the  detent  d  from  its  support  r,  thus  allowing  the  tooth  b  to  escape, 
the  escape-wheel  being  urged  in  the  direction  of  the  arrow.  While  this 
is  occurring,  the  tooth  62  encounters  the  projection  n,  and  gives  an 
impulse  to  the  balance;  the  detent  meanwhile  has  dropped  back  under 
the  influence  of  the  spring  s,  and  catches  the  next  tooth  of  the  wheel  61. 

It  will  be  noticed  that  the  impulse  is  given  to  the  balance  immediately 
after  it  has  been  subject  to  the  resistance  of  unlocking  the  detent  d, 
thus  immediately  compensating  this  resistance;  also  that  the  impulse 
is  given  at  every  alternate  swing  of  the  balance. 

The  motion  of  the  balance  is  so  adjusted  that  the  impulse  is  given 
through  equal  distances  on  each  side  of  the  middle  of  its  swing. 


PROBLEMS 

1.  Referring  to  Fig.  1,  page  7,  if  OR  is  a  crank  18  ins.  long  turning  90  r.p.m. 
find  the  displacement  and  acceleration  of  point  T  when  6  =  60  degrees. 

2.  If  the  periphery  speed  of  a  drill  2  ins.  in  diameter  is  not  to  exceed  40  ft.  per 
minute  what  is  the  maximum  number  of  r.p.m.  at  which  it  may  be  run? 


M 


3.  (a)  What  is  the  angular  speed  of  this 
pulley  in  radians  per  second? 

(6)  What  is  the  linear  speed  of  a  point  M  on 
the  circumference  in  feet  per  minute? 

(c)  What  is  the  ratio  of  the  linear  speed  of 
P  to  that  of  M? 


PROB.  3 

4.  Assume  that  in  turning  steel  shafting  a  cutting  speed  of  80  ft.  per  minute 
allowable.     How  fast  must  a  piece  of  6-in.  shaft  turn  to  give  this  speed? 


6.  The  Pelton  water  wheel  shown  ia  the  figure 
is  driven  by  a  jet  of  water  having  a  velocity  of 
90  feet  per  second.  The  linear  speed  of  the 
center  line  of  the  buckets  is  0.6  that  of  the 
water.  What  is  the  diameter  in  inches  to  the 
center  line  of  the  buckets  if  the  wheel  is  fast  to 
a  shaft  that  turns  100  r.p.m.? 


6.  This  emery  wheel  is  6  ins.  in  diameter  and  makes 
2546  r.p.m. 

(a)  What  is  its  surface  speed  in  ft.  per  minute? 

(6)  If  the  wheel  turns  as  indicated  and  the  piece  of  cast 
iron  D  is  fed  as  shown  by  the  arrow  at  the  rate  of  30  ft. 
per  minute,  what  is  the  cutting  speed  ? 

(c)  If  D  is  fed  in  the  same  direction  and  at  the  same 
speed  as  before  and  if  the  emery  wheel  turns  2546  r.p.m. 
but  in  a  clockwise  direction,  what  is  the  cutting  speed  ? 

336 


PBOB.  5 


PROB.  6 


PROBLEMS 


337 


7.  A  pulley  makes  400  r.p.m.    A  point  on  its  outer  surface  has  a  linear  speed 
of  4000  ft.  per  minute.    What  is  the  diameter? 


8.  The  sketch  shows  a  grindstone 
operated  by  a  treadle.  If  IT  is  taken 
as  3  and  the  average  speed  of  the  sur- 
face of  the  stone  is  540  ft.  per  minute, 
through  what  distance  does  the  point 
F  on  the  treadle  move  in  15  minutes? 
Through  how  many  radians  does  it 
move  in  that  time? 

Note.  —  The  point  E  is  on  the  verti- 
cal center  line  YY  when  the  treadle  is 
in  the  highest  and  lowest  positions. 


Fulcrum 


PROB.  8 

9.  On  a  revolving  wheel  are  two  points  located  on  the  same  radial  line.     If  the 
wheel  has  an  angular  speed  of  1000  radians  per  minute  and  if  the  linear  speed  of 
one  point  is  3500  ins.  per  minute  greater  than  the  other,  how  far  apart  are  the 
points  ? 

10.  The  shaft  of  a  centrifugal  drying  machine  has  an  angular  speed  of  4398.24 
radians  per  minute.     What  is  the  linear  speed  of  a  point  2  ft.  out  from  the  center 
of  the  shaft  in  ft.  per  minute  and  how  many  r.p.m.  does  the  dryer  make? 

11.  A  bell-crank  lever  of  the  type  shown  in  Fig.  17  has  arms  that  are  8  and  10 
ins.  long  and  make  an  angle  of  60  degrees  with  each  other.     If  the  long  arm  is  hori- 
zontal and  has  a  100-lb.  weight  suspended  from  its  end,  what  vertical  pull  must  be 
exerted  at  the  end  of  the  short  arm  to  balance  the  100-lb.  weight? 

12.  In  a  rocker  of  the  type  shown  in  Fig.  18  the  line  of  motion  of  the  end  of  one 
arm  is  vertical  while  the  line  of  motion  of  the  end  of  the  other  is  45  degrees  with  the 
horizontal.    One  arm  is  5  ins.  long  and  the  motion  of  its  end  along  the  horizontal 
line  is  2 1  ins.,  the  motion  of  the  end  of  the  other  arm  is  1^  ins.  along  the  45-degree 
line.     If  the  ends  of  the  arms  of  the  bell  crank  swing  equally  either  side  of  the 
lines  of  motion,  locate  graphically  the  fulcrum  of  the  rocker  and  show  the  center 
lines  of  the  arms  when  in  their  mid-positions. 

13.  The  main  driving  pulley  of  a  broaching  machine  is  18  ins.  in  diameter  and  turns 
at  a  speed  of  400  r.p.m.     If  the  pulley  is  driven  by  a  belt  from  a  10-H.P.  motor 
developing  its  full  rated  power,  what  is  the  effective  pull  of  the  belt  ? 

14.  A  shaft  making  200  r.p.m.  carries  a  pulley  36  ins.  in  diameter  driven  by  a  belt 
which  transmits  5  horse  power.     What  is  the  effective  pull  in  the  belt?    What 
should  be  the  diameter  of  the  pulley  if  the  effective  pull  is  to  be  50  Ibs.  ? 

15.  A  6-in.  belt  transmits  8  horse  power,  when  running  over  a  pulley  20  ins.  in 
diameter  running  200  r.p.m.    If  the  maximum  allowable  tension  is  80  Ibs.  per  inch 
of  width  and  if  the  sum  of  the  tensions  is  assumed  to  be  constant,  with  what  tension 
was  the  belt  put  on  the  pulleys? 


338  PROBLEMS 

16.  What  is  the  width  of  a  fabric  belt  which  transmits  45  horse  power  when  run- 
ning on  a  30-in.  pulley  which  turns  550  r.p.m.  ?    The  belt  was  put  on  with  a  tension 
of  70  Ibs.  per  inch,  the  maximum  tension  is  not  to  exceed  95  Ibs.  per  inch,  neglecting 
centrifugal  force,  and  it  is  assumed  that  the  sum  of  the  tensions  is  a  constant. 

17.  Determine  the  width  of  a  single  belt  to  carry  40  horse  power  when  running 
on  a  pulley  48  ins.  in  diameter  which  turns  300  r.p.m.     The  maximum  tension  per 
inch  of  width  for  a  single  belt  is  65  Ibs.,  neglecting  centrifugal  force,  and  it  is  assumed 
that  Ti  =  2£  T2.     Determine  the  width  of  this  belt,  using  the  millwrights'  rule. 

18.  A  3-in.  belt  is  designed  to  stand  a  difference  in  tension  of  50  Ibs.  per  inch  of 
width,  neglecting  centrifugal  force.    Find  the  least  speed  at  which  it  can  be  driven 
in  order  to  transmit  20  horse  power. 

19.  To  what  difference  in  tension  on  the  two  sides  of  a  belt  does  the  millwrights' 
rule  correspond?    If  the  maximum  tension  for  a  single  belt  is  65  Ibs.  per  inch  of 

T 

width,  and  for  a  double  belt  150  Ibs.  per  inch  of  width,  what  is  the  value  of  -^  in  a 

I* 

belt  calculated  by  the  millwrights'  rule?    Answer  both  parts  of  the  problem  for 
both  single  and  double  belt,  neglecting  the  effect  of  centrifugal  force. 

20.  Shaft  A,  turning  120  r.p.m.,  drives  shaft  B  at  speeds  of  80, 120, 180  and  240 
r.p.m.  by  means  of  a  pair  of  stepped  pulleys  and  a  crossed  belt.     The  diameter  of 
the  largest  step  on  the  driving  pulley  is  18  ins.     Calculate  the  diameters  of  all  the 
steps  of  both  pulleys. 

21.  Two  shafts,  each  carrying  a  four-step  pulley,  are  to  be  connected  by  a  crossed 
belt.     The  driving  shaft  is  to  turn  150  r.p.m.  while  the  driven  shaft  is  to  turn  50, 
150,  250  and  600  r.p.m.     The  smallest  step  of  the  driver  is  10  ins.  in  diameter. 
Find  the  diameters  of  all  the  steps. 

22.  Solve  the  preceding  problem  if  an  open  belt  is  used  instead  of  a  crossed 
belt  and  if  the  shafts  are  30  ins.  apart. 

23.  Two  shafts  carrying  five-step  pulleys  are  to  be  connected  by  a  crossed  belt. 
'The  driver  is  to  turn  150  r.p.m.  while  the  follower  is  to  have  speeds  of  50,  100,  150, 

200  and  250  r.p.m.     If  the  smallest  step  on  either  pulley  is  8  ins.  in  diameter,  find 
the  diameters  of  all  the  steps  to  two  decimal  places. 

24.  The  feed  mechanism  of  an  upright  drill  is  operated  by  an  open  belt  running 
on  three-step  pulleys.     The  driving  shaft  turns  150  r.p.m.  while  the  driven  shaft 
turns  150,  450  and  900  r.p.m.,  the  two  shafts  being  15  ins.  apart.     If  the  largest 
diameter  of  the  driver  is  18  ins.  find  the  diameters  of  all  the  steps.     If  the  steps  of 
these  pulleys  had  been  calculated  for  a  crossed  belt  but  an  open  belt  had  been  used 
on  them  the  belt  would  have  been  found  too  short  to  run  on  some  of  the  steps. 
State  approximately  how  much  too  short  it  would  have  been  for  the  worst  case. 

25.  A  lathe  having  a  five-step  pulley  is  driven  by  a  belt  (assumed  to  be  crossed) 
from  a  pulley  of  the  same  size  on  the  countershaft.     The  countershaft  is  to  have  a 
constant  speed,  and  the  lathe  is  to  have  speeds  of  60  r.p.m.  and  135  r.p.m.  when  the 
belt  is  on  the  steps  either  side  of  the  center  step.    If  the  minimum  speed  is  40  r.p.m. 
and  the  smallest  diameter  4  ins.,  find  proper  speed  of  countershaft,  maximum  speed 
of  lathe,  and  diameter  of  all  steps  on  the  pulleys. 

26.  Each  of  a  pair  of  equal  five-stepped  pulleys  has  diameters  20,  17f,  15J,  12|, 
10|  ins-    Were  these  pulleys  designed  for  an  open  belt  or  a  crossed  belt?    If  the 
driving  shaft  turns  500  r.p.m.  calculate  the  five  speeds  of  the  driven  shaft. 

27.  In  a  pair  of  stepped  pulleys  the  driver  has  diameters  of  31.62,  25.5,  20.53,  10 
ins.    The  smallest  diameter  of  the  driven  pulley  is  7.91  ins.  and  its  largest  diam- 
eter 30  ins.    Is  the  belt  crossed  or  open  ?    Calculate  the  distance  between  centers 
of  the  shafts  and  the  other  two  diameters  of  the  driven  pulley. 


PROBLEMS 


339 


28.  Two  shafts  are  connected  by  a  crossed  belt  running  on  a  pair  of  speed  cones. 
The  driving  shaft  has  a  constant  speed  of  135  r.p.m.  while  the  driven  shaft  is  to 
have  a  range  of  speeds  from  45  to  300  r.p.m.,  the  speeds  to  increase  in  arithmetical 
progression  as  the  belt  is  moved  equal  distances  along  the  cones.     The  smallest 
diameter  of  the  driving  cone  is  3  ins.    Find  the  diameters  of  the  cones  at  the  ends 
and  at  two  intermediate  points.     Plot  the  cones  (|  size)  if  their  length  is  24  ins. 

29.  A  rope  drive  composed  of  15  ropes  is  to  transmit  120  horse  power.     If  the 
pitch  diameter  of  one  of  the  sheaves  is  4  ft.  and  if  its  angular  speed  is  1500  radians 
per  minute,  what  is  the  effective  pull  in  each  rope  ? 

30.  A  rope  drive  (Multiple  System)  consisting  of  15  ropes  is  transmitting  200 
horse  power  when  the  speed  of  the  ropes  is  1100  ft.  per  minute.     The  maximum 
tension  per  rope  is  650  Ibs.  which  is  |  the  breaking  strength  of  the  rope  (expressed 

rri 

as,  "a  factor  of  safety  of  4").     Find  the  ratio  ^-     Suppose  that  3  of  the  ropes 

2  Ti 

should  break,  the  remaining  ropes  carrying  the  whole  load.     If  the  ratio  ^-  stays 

as  before,  what  does  the  maximum  tension  become  and  what  the  factor  of  safety 
on  the  rope? 

31.  A  motor  running  500  r.p.m.  transmits  3  horse  power  through  a  chain  drive. 
The  pitch  diameter  of  the  driving  sprocket  is  3  ins.     What  is  the  effective  pull  in 
the  chain  and  what  is  the  maximum  tension  neglecting  centrifugal  force? 

32.  A  chain  drive  is  transmitting  3  horse  power  when  the  speed  of  the  chain  is 
900  ft.  per  minute.     What  is  the  effective  pull  in  the  chain?    Suppose  the  speed  of 
the  chain  be  increased  to  1200  ft.  per  minute,  if  the  power  transmitted  remains  the 
same  as  before,  what  is  the  effective  pull? 

75  R.P.M. 


33.  How  many  r.p.m.  does  B  make? 


34.  Angular  speed  of  S  =  |  of  the  ang- 
ular speed  of  T.  Calculate  and  find  gra- 
phically the  diameters  of  cylinders  to 
connect  them: 

(a)  When  they  turn  as  shown  by  the 
full  arrows. 

(6)  When  they  turn  as  shown  by  the 
dotted  arrows. 


PROB.  33 


PROB.  34 


36.  A  and  B  are  rolling  cylinders  connecting 
the  shafts  S  and  T.  C  and  E  are  cylinders  fast 
to  these  shafts  and  slipping  on  each  other  at  P. 
Find  the  diameters  of  C  and  E  if  the  surface  speed 
of  E  is  twice  that  of  C. 


C 

x 

E 

H 

~ 

1     H 

B 

16  D.- 
PROB.  35 


340 


PROBLEMS 


36.  The  shaft  of  a  grindstone  is  mounted  on 
rollers,  as  shown.  If  the  circumference  of  the 
stone  has  a  speed  of  628.32  ft.  per  minute,  find 
its  r.p.m.  If  the  angular  speed  of  the  rollers  is 
209.44  radians  per  minute,  find  their  diameter. 


PROB.  36 


37.  How  far  from  the  axis  of  T  will  the 
center  of  the  roller  R  be  located  if  the 
angular  speed  of  shaft  S  is  three  times  as 
great  as  that  of  T? 


PROB.  37 


38.  A  and  B  are  two  shafts  at  right  angles,  in  the  same 
vertical  plane.  C  is  a  disk  carried  by  a  supporting  yoke 
on  a  horizontal  shaft  arranged  so  that  C  is  always  in 
contact  with  the  equal  conoids  on  A  and  B.  A  turns  at  a 
constant  speed  of  60  r.p.m.  What  is  the  maximum  speed 
of  Bl  What  is  the  minimum  speed  of  B?  What  is  the 
speed  of  B  when  the  yoke  supporting  C  has  turned  30 
degrees  from  its  present  position  ?  (Assume  no  slipping.) 


PROB.  38 


39.  A  turns  100  and  B  150 
r.p.m.  as  shown  and  are  connected 
by  rolling  cones.  Calculate  the 
apex  angle  of  each  cone.  If  the 
base  of  cone  on  A  is  3  ins.  from 
the  vertex,  calculate  the  diameters 
of  both  cones.  Solve  also  graphi- 
cally 


PROB.  39 


PROB.  40 


40.  Two  shafts  A  and  B  are  connected  by  rolling  cones  and  turn  as  shown.  A 
makes  300  r.p.m.  while  B  makes  100  r.p.m.  Calculate  the  apex  angle  of  each  cone 
and  the  diameter  of  each  base  if  the  base  of  cone  B  is  2  ins.  from  the  vertex.  Solve 
also  graphically. 


PROBLEMS 


341 


41.  Shaft  S  makes  180  r.p.m.  and  shaft  T  makes  60  r.p.m.  Draw  a  pair  of 
frustra  of  cones  to  connect  them.  Base  of  smaller  cone  1  in.  in  diameter.  Element 
of  contact  1  in.  long. 

(a)  When  the  shafts  turn  as  shown  by  the  full  arrows. 

(6)  When  they  turn  as  shown  by  the  dotted  arrows. 


PROB.  41 


42.  Shafts  A,  B  and  C  are  connected  by  cones  in  external  rolling  contact  so 
that  the  revolutions  A  :  B  :  C  =  3  :  2  :  4.     If  the  diameter  of  cone  B  is  6  ins. 
draw  in  the  three  cones  giving  the  diameters  of  cones  A  and  C.     (Show  method 
clearly.) 

43.  Given  a  4-pitch  gear  of  24  teeth.    The  addendum  equals  the  module,  the  clear- 
ance is  to  be  |  of  the  addendum  and  the  back  lash  is  to  be  2  per  cent  of  the  circular 
pitch.     Calculate  the  following,  giving  results  to  three  decimal  places:  the  pitch 
diameter,  the  diameter  of  the  blank  gear  before  cutting  the  teeth  (addendum  diam- 
eter), the  depth  of  teeth,  the  backlash  and  the  width  of  tooth  and  width  of  space. 


44.  In  the  given  position 

angular  speed  of  A 
angular  speed  of  B 

is  equal  to  the  angular  speed  ratio  of  what 
two  rolling  cylinders?  Prove  that  this  is  so. 
(Make  diagram  full  size.) 


PROB.  44 


45.  A  pinion  6-ins.  in  pitch  diameter  is  to 
drive  a  rack.  The  flank  of  the  rack  is  an  arc 
of  a  circle  2|-ins.  in  radius  with  its  center  1 1  ins. 
below  pitch  line  of  rack  as  shown.  Find  two 
points  on  the  face  of  the  pinion  starting  with 
points  on  given  flank  \  and  \  in.  from  the 
pitch  point.  Assuming  an  addendum  of  \  in. 
on  pinion  show  arc  of  recess  on  the  rack's 
pitch  line. 


PROB.  45 


342 


PROBLEMS 


46.  Given  this  flank  of  a  rack,  a  straight  line  at  15 
with  line  of  centers,  find  two  points  on  the  face 
of  the  pinion  about  j  and  \  in.  away  from  the  pitch  line, 
and  show  path  of  contact  in  recess  if  the  pinion  drives 
turning  right-handed. 


PROB.  46 


47.  A  pinion  5  ins.  in  diameter  is  to  drive 
an  annular  12  ins.  in  diameter.  The  flanks 
of  the  annular's  teeth  are  arcs  of  circles  with 
2-in.  radius  located  as  shown.  Find  three 
points,  as  a,  b,  c,  on  the  face  of  the  pinion. 
Show  the  path  of  contact  in  recess  if  the 
pinion's  addendum  is  1  in.  Find  the  arc  of 
recess  and  the  angles  of  recess,  and  the  length 
of  the  acting  flank. 


PROB.  47 

48.  A  pinion  10-ins.  in  diameter  with  radial  flanks  is  to  be  driven;by  a  pinion  4-in. 
diameter  also  having  radial  flank  teeth. 

1°  Find  the  respective  faces. 

2°  If  the  addendums  =  \  in.  show  the  path  of  contact;  the  angles  of  approach 
and  recess;  the  maximum  angles  of  obliquity  in  approach  and  recess;  and  the 
lengths  of  the  acting  flanks. 

3°  Draw  the  true  clearing  curve  which  the  4-in.  pinion's  tooth  would  trace 


49.  Diameter  of  pitch  circle  of  pinion 
=  3£  ins.  Diameter  of  pitch  circle  of  an- 
nular =X8  ins.  The  flank  of  the  annular  is 
to  be  the  arc  of  a  circle  1$  ins.  in  radius 
as  shown.  Find  two  points  on  the  face 
of  the  pinion  starting  with  points  on  given 
flank  |  and  f  in.  from  the  pitch  point. 
Show  path  of  contact  in  approach  if  annular 
drives  left-handed. 


PROB.  49 


PROBLEMS  343 

60.  Involute  gears,  22^  degrees  obliquity,  1-pitch,  addendum  =  f  the  module, 
clearance  =  |  the  module,  no  backlash.  A  pinion  having  9  teeth,  turning  clockwise, 
is  to  drive  a  gear  of  12  teeth.  Indicate  the  path  of  contact,  and  angles  of  approach 
and  recess  for  each  gear,  also  give  the  ratio  of  the  arc  of  action  to  the  circular  pitch. 
Draw  two  teeth  on  each  gear,  having  a  pair  of  teeth  in  contact  at  the  pitch  point. 
Indicate  the  acting  flank  of  the  teeth  on  each  gear. 

51.  Find  the  diameter  and  number  of  teeth  of  the  smallest  3-pitch  pinion  with 
20  degrees  obliquity,  which  would  allow  an  arc  of  recess  =  arc  of  approach  =  the 
circular  pitch,  and  draw  its  pitch  and  addendum  circles.     If  the  pinion  drives  a  rack 
what  is  the  greatest  allowable  addendum  for  the  rack? 

52.  Involute  gears,  15  degrees  obliquity;   a  30-tooth  pinion  2-pitch  is  to  drive  a 
rack.     How  long  can  the  arc  of  approach  be?     Can  the  arc  of  recess  equal  the  circu- 
lar pitch  and  why  ? 

53.  An  involute  gear  with  21  teeth,  3-pitch,  15  degrees  obliquity,  has  an  adden- 
dum diameter  of  1\  ins.     Draw  its  base  circle  and  pitch  circle.     Could  two  such 
gears  properly  be  used  to  connect  two  shafts  7|  ins.  apart?    Give  reason  for  answer. 

54.  Involute  gears,  2-pitch,  15  degrees  obliquity.     A  24-tooth  pinion  is  to  drive  a 
32-tooth  annular.    The  arc  of  approach  to  be  equal  to  1^  the  pitch  and  arc  of  recess 
to  be  equal  to  the  circular  pitch.     Draw  the  addendum  circles.     Is  each  of  these  arcs 
possible;  (explain  clearly  the  steps  by  which  this  is  determined).     What  is  the  limit 
of  the  path  of  contact  in  approach  and  in  recess  and  why? 

55.  Two  cycloidal  gears  with  10  and  12  teeth  respectively;  2-pitch;  radial  flanks 
on  each. 

1°  Draw  the  pitch  circles  and  the  describing  circles  and  give  their  diameters. 

2°  If  the  addendum  =  \  in.  and  the  10-tooth  pinion  drives  turning  right  handed 
show  the  path  of  contact. 

3°  How  long  is  the  arc  of  action  in  terms  of  the  pitch  ?  How  long  must  it  be  to 
just  give  perfect  action? 

56.  A  pinion  with  6  teeth  1-pitch  is  to  drive  one  with  8  teeth.     Radial  flanks  on  8- 
tooth  and  the  same  size  describing  circle  for  the  flanks  of  the  6-tooth.     The  arc 
of  approach  to  be  f  of  the  pitch  and  the  arc  of  recess  to  be  f  of  the  pitch. 

1°  Find  the  maximum  angles  of  obliquity  in  approach  and  in  recess  in  degrees. 
2°  Is  the  given  arc  of  action  possible? 

67.  Cycloidal  Gears.     Interchangeable  Set.     3-pitch.     Radial  flanks  on  a  15- 
tooth  gear.     Addendum  equals  module.     Clearance  equals   |  of  the  addendum. 
An  18-tooth  pinion  drives  a  39-tooth  annular.     Show  path  of  contact.     How  many 
teeth  would  there  be  in  the  smallest  annular  that  would  gear  with  the  18-tooth  pinion  ? 
Show  path  of  contact  in  this  case. 

68.  In  a  f-pitch,  interchangeable  set  of  cycloidal  gears  with  addendum  the  same 
on  all  gears,  it  is  found  that  two  8-tooth  pinions  will  give  a  path  of  contact  2 
inches  long.     Could  one  of  the  8-tooth  pinions  properly  drive  a  7-tooth  pinion  of 
the  set? 

59.  A  pin-gear  1-pitch  with  8  pins  is  to  be  driven  by  a  rack.     The  pins  are  to  be 
one-half  the  pitch  in  diameter,  and  the  addendum  on  the  rack's  teeth  is  1^  ins. 
Find  the  true  path  of  contact.     Also  draw  the  teeth  for  the  rack  and  the  pins  for 
the  gear,  assuming  no  back  lash,  and  a  clearance  of  0.1  in. 

60.  A  6-tooth  1-pitch  cycloidal  gear  has  its  flank  describing  circle  equal  to  the 
pitch  circle.     It  is  to  drive  another  6-tooth  gear  like  itself.     Draw  the  addendum 
circles  so  that  the  arcs  of  approach  and  recess  are  each  equal  to  f  of  the  pitch,  and 
show  the  path  of  contact.     What  would  be  the  shapes  of  the  teeth? 


344 


PROBLEMS 


61.  Shaft  A  turns  120  r.p.m.  in  the  di- 
rection shown  and  drives  shaft  B  by  means 
of  an  open  belt  running  on  the  right-hand 
steps  of  the  pulleys.     Shaft   C  is   driven 
from  B  by  a  pair  of  gears  so  that  C  turns 
3  times  for  every  2  turns  of  B.     Gear  D 
has  26  teeth  while  E  has  78  teeth.     Shaft 
F  carries  a  bevel  gear  of  12  teeth  which 
drives  one  of  120  teeth  on  shaft  G.    Shaft 
G  also  carries  H,  a  4-pitch,  16-tooth  gear 
which  is  in  mesh  with  a  sliding  rack.     What 
is  the  speed  of  the  rack  in  inches  per  minute 

and  does  it  move  to  the  right  or  left?  PROB.  61 

62.  In  a  broaching  machine,  the  shaft  A  carries  a  pulley 
24  ins.  in  diameter  which  is  driven  by  a  belt  from  a  12-in. 
pulley  on  the  countershaft  overhead,  the  latter  turning 
150  r.p.m.     The  gears  B  and  D  have  12  teeth  each,  while 
C  and  E  have  60  teeth.    Gear  E  is  fast  to  F,  which  has 
10  teeth  and  a  circular  pitch  of   1.047  ins.  and  which 
engages  with  rack  G  to  which  is  attached  the  broach. 
Find  the  speed  with  which  the  broach  is  drawn  through 

the  work  in  inches  per  minute.  PROB.  62 

63.  In  a  brick-making  machine  is 
found  this  train  of  gears.     A  motor 
carrying  pulley  E,  which  is  6  ins.  in 
diameter,  drives  the  machine.     The 
wide-faced  roller  F,  12  ins.  in  diam- 
eter, drives  a  conveyor  belt.    If  the 
motor  runs  at  1200  r.p.m.  what  is  the 
speed  of  the  conveyor  belt  in  ft.  per 
minute?     (Neglect  the  thickness  of 

the  belt.)  PROB.  63 

64.  In  a  crane,  the  chain  barrel  is  driven  by  a  motor  on  the  spindle  of  which  is 
keyed  a  pinion  of  14  teeth.     This  gears  with  a  wheel  of  68  teeth  keyed  to  the  same 
spindle  as  a  pinion  of  12  teeth.     The  last  wheel  gears  with  a  wheel  of  50  teeth  keyed 
to  the  same  spindle  as  a  wheel  of  25  teeth,  and  the  latter  gears  with  a  wheel  of  54 
teeth  keyed  to  the  chain  barrel  spindle.    Chain  barrel  is  16|-ins.  in  pitch  diameter. 
Sketch  the  arrangement  and  find  r.p.m.  of  motor  when  20  ft.  of  chain  are  wound 
on  drum  per  minute. 


65.  Effective  pull  on  the  belt  is  250  Ibs. 
W=  7000  Ibs.  A  is  23  ins.  in  diameter.  B, 
D,  and  F  each  have  18  teeth.  E  =  63  teeth; 
G  =  50  teeth.  H  is  20  inches  in  pitch  diam- 
eter. If  there  is  a  loss  of  power  of  60  per  cent, 
how  many  teeth  must  there  be  in  gear  C? 
Which  is  the  tight  side  of  the  belt? 


PROB.  65 


PROBLEMS 


345 


66.  Sketch  shows  side  elevation  of  a 
molding  machine.  The  stock  is  fed  through 
rolls  A  to  cutter  C  which  is  driven  by  a 
quarter  turn  belt  as  shown.  Rolls  A  are 
4^  ins.  in  diameter,  the  upper  one  only  being 
power  driven.  If  the  cutter  C  is  6  ins.  in 
diameter,  find  the  feed  of  the  stock  per 
revolution  of  cutter  and  the  relative  speed 
of  cutter  and  work.  (Cutter  is  shown  be- 
hind the  work.) 


•12  T. 


PROB.  66 


67.  A  is  an  annular  gear  having  77  teeth, 
driving  pinion  B  having  12  teeth.  Numbers 
of  teeth  on  the  other  gears  as  are  given  in  the 
figure.  If  A  makes  15  r.p.m.  find  the  rate  of 
slip  between  the  cylinders  R  and  S  in  feet  per 
second. 


PROB.  67 


68.  The  back  gears  of  an  engine  lathe  train  are  to  give  a  reduction  in  the  ratio 
of  f .  Arrange  a  train  to  give  this,  using  no  wheel  of  less  than  15  teeth.  First  pair 
of  gears  4-pitch,  second  pair  3-pitch.  Make  reduction  of  speed  by  the  two  pairs  as 
nearly  equal  as  possible. 


69. 


Rev.  B      19 


T:-;  find  suitable  numbers  of  teeth  for  the  four 


Rev.  A 

gears  of  this  train,  having  all  of  the  same  pitch, 
have  more  than  75  teeth  nor  less  than  10  teeth. 


No  gear  to 


PROB.  69 


70.  Shaft  S  has  a  constant  speed  of  100  r.p.m. 
Gears  F,  G,  and  H  form  a  unit  free  to  slide,  but 
not  to  turn  on  shaft  Si.  Si  is  to  have  speeds  of 
20,  200  and  860  r.p.m.  Gear  H  has  80  teeth. 
Find  numbers  of  teeth  on  all  gears  if  they  are 
all  of  the  same  pitch,  and  if  gears  A  and  B  are 
equal.  The  slowest  speed  of  Si  is  when  E  and 
H  are  in  mesh. 


PROB.  70 


346 


PROBLEMS 


71.  The  sketch  shows  a  nest  of  spur  gears,  each  pa^r  being 
always  in  mesh  and  all  gea^s  of  the  same  pitch.  F,  G,  and  H 
form  a  unit  keyed  to  shaft  B.  C,  D,  and  E  are  loose  on  shaft  A, 
but  may  be  locked  to  the  shaft  one  at  a  time.  Find  the  num- 
bers of  teeth  on  all  gears  if  the  revolutions  of  B  for  one  of  A  are 
to  be  £,  If,  and  2£  as  C,  D  and  E  respectively  are  keyed  to  A. 
No  gear  to  have  less  than  14  teeth. 


JH 

G 


PROB.  71 


72.  Both  gears  on  both  shafts  A  and  B  are  fast. 
The  three  gears  on  shaft  C  are  fast  together  but  not 
to  the  shaft  and  may  be  moved  along  shaft  C  as  a 
unit.  Shaft  A  has  a  constant  speed  of  60  r.p.m.  and 
speeds  of  B  are  to  be  240,  60  and  15  r.p.m.  Find 
suitable  numbers  of  teeth  in  all  gears  if  they  are  all 
4-pitch. 


PROB.  72 

73.  The  value  of  a  train  is  to  be  approximately  113,  with  an  allowable  variation 
of  1.    Find  the  least  number  of  pairs  and  sketch  the  train,  indicating  the  numbers 

f  teeth  on  each  gear.     No  gear  to  have  more  than  50  teeth  nor  less  than  12  teeth. 

T. 

74.  A,  B,  and  C  are  gears  having  teeth 

as  shown.  /  \   /  \    29  r. 

1°  If  A  turns  +4  and  arm.  turns  -3, 
find  turns  of  B  and  C. 

2°  If  A  is  not  to  turn  and  B  turns  +30, 
find  turns  of  the  arm. 


75.  Gear  A  is  fixed.  The 
arm  turns  about  the  shaft  on 
which  A  is  located.  Find  the 
number  of  teeth  on  -gear  C  if 
it  makes  three  times  as  many 
absolute  turns  as  B  does  but 
in  the  opposite  direction. 


40T, 


20  T, 


35  T, 


PROB.  75 


PROBLEMS 


347 


76.  In  this  roller  bearing  D  represents 
the  fixed  bearing  in  which  the  rollers  are 
supported  while  S  is  the  shaft.  Assuming 
that  there  is  pure  rolling  contact  between 
the  shaft  and  the  rollers,  and  between  the 
rollers  and  D,  find  the  ratio  of  speed  at 
which  roller  cage  revolves  to  speed  at  which 
shaft  revolves. 


77.  1°  If  A  turns  +3 
and  the  arm  —5,  find  the 
turns  of  B,  C,  and  D. 

2°  Suppose  two  idlers  to 
be  used  between  E  and  D, 
other  conditions  remaining 
as  before,  find  the  turns  of 
D. 


50  T. 


PROB.  76 


eor, 


60  T, 


25  T. 


PROB.  77 


78.   If  A  turns  +38  times,  how 
many  turns  does  the  arm  make  ? 


Fixed 


PROB.  78 


348 


PROBLEMS 


120  T. 


79.  If  shaft  A  makes  20  turns  in  a  positive 
direction,  find  the  number  of  turns  of  shaft  B 
and  its  direction. 


32  T, 


Area 


60  T, 


'"    20  T, 


80  T, 


a 7.  SOT. 


PROS.  79 


80  T. 


80.  D  is  a  fixed  gear,  A  is  fast  to  the 
shaft  with  the  54-toothed  gear,  and  P  is  the 
arm  of  the  epicyclic  train.  If  A  turns  -20 
find  the  turns  of  the  arm,  then  find  the  turns 
of  B. 


8( 

1 

r, 

2 

)  1 

i_ 

"™"""" 

i  — 

- 

n; 

P 

— 

- 

36  1 



- 

— 

- 

- 



A         D 

s 

/ 

r. 

5 

4 

r, 

8 

1  1 

48  T. 


PROS.  80 


81.  In  this  hoisting  mechanism,  A  is 
a  fixed  annular  having  100  teeth.  The 
two  idle  pinions  B  are  carried  by  the  arm 
of  the  epicyclic  train,  which  also  carries 
the  drum,  as  shown.  Gear  C,  which  is 
fast  to  the  crank,  has  70  teeth.  Diameter 
of  drum  is  5  ins.,  length  of  crank  is  21 
ins.,  force  applied  at  crank  is  75  Ibs. 
Find  the  teeth  on  the  pinions  B  and 
weight  lifted,  neglecting  friction. 


PROS.  81 


PROBLEMS 


349 


30  T, 


82.  For  36  turns  of  D,  find  how        20 
many  turns  of  F  and  in  which  direction  ? 

23  r. 


83.  For  -3  turns  of  D, 
find  how  many  of  F  and  in 
which  direction? 


\ 

i 

— 

- 

'~ 

r 

- 

— 

J 

40  1 

^ 

r 

g 

\ 

— 

T 

—  —  ' 

5 

0 

30 

r, 

L 

20  T 

'\ 

PI 

ft 

?  7 

^ 

^ 

C 

/ 

S  7 

-T 

1 

/«  7 

PBOB.  82 


20  Ti 

/} 

C 

D 

/ 

5 

r, 

*~40  T, 

— 

36  1 

c 

13 



/ 

5' 

'. 

20  : 

45  T. 


20 


W  T. 


flOO 


PROS.  83 


~30  T. 


30  T. 


1ST. 


84.  In  this  train  find  the 
ratio  of  the  diameters  C  and 
D,  if  3  turns  of  A  as  shown 
are  to  cause  the  arm  to  turn 
11  times.  Must  a  crossed 
or  an  open  belt  be  used  if 
the  arm  turns  as  shown  ? 


16 T 


=_.. 


55  r. 


PBOB.  84 


80  T, 


LJ 


350 


PROBLEMS 


25  T. 


86.   Let  shaft  S  turn  +3  times.    Find 
the  turns  of  the  arm. 


307. 


86.  If  A  is  a  shaft  coupled 
to  a  dynamo  making  2500 
r.p.m.,  how  many  revolutions 
per  minute  does  B  make? 


PROB 


18  T. 

60  T.  \. — ,       30  T. 


50  T. 


17  T, 

87.  The  gear  B  is  fixed.  For  31 
turns  of  A  hi  direction  shown,  how 
many  turns  does  C  make  and  does 
it  turn  in  the  same  direction  as  A 
or  opposite? 


14  T. 


24  T. 


42T. 


PROB.  87 


PROBLEMS 


351 


50  T, 


89  T. 


88.   In  this  epicyclic  train  let  A 
turn  +6,  find  the  turns  of  B. 


40  T, 


90  T. 


80  T. 


90  T, 


60  T. 


PROB.  88 


89.  The  pulley  P  makes  40  r.p.m.  in  the  direction  shown. 
What  must  be  the  lead  of  the  screw  if  nut  A  is  to  rise  3f  ins. 
in  45  seconds?  Is  screw  right  handed  or  left  handed? 


90.  What  pressure  in  Ibs.  on  the  square  inch  is  exerted  on 
a  liquid  below  the  piston  by  a  force  of  40  Ibs.  at  the  rim  of 
the  hand-wheel?  The  screw  has  i-in.  lead  and  is  double 
threaded.  Allow  20  per  cent  lost  in  friction. 


UUb 


PROB.  89 


PROB.  90 


91.  Find  lead  in  inches  of  the  screw  if  a  force 
of  If  tons  is  to  be  exerted  at  W  by  a  pull  of  75 
pounds  on  the  rope  in  the  groove  of  the  wheel  D 
which  has  an  effective  diameter  of  4|  ft.  (Allow 
45  per  cent  friction  loss.) 


PROB.  91 


352 


PROBLEMS 


92.  If  driving  pulley,  D,  makes  300  r.p.m. 
at  what  rate  is  the  cross  rail  raised? 


;'/4  Lead  R.  H^ 


93.  B  and  C  are  two  equal  gears.  They 
may  have  no  axial  motion.  D  and  E  are 
two  gears,  E  being  four  times  .as  large  as  D. 
Shaft  A  is  a  square  shaft  turning  with  B,  but 
free  to  slide  through  it.  The  screw  threaded 
through  C  has  a  lead  of  |  in.  left  handed. 
How  many  turns  of  the  handle  are  needed 
and  which  way  (front  side  of  D  going  down 
or  up)  to  move  the  screw  2  ins.  to  the  right? 


94.  20  turns  of  F  are  to  raise  W  5|  ins.  Pi  =  0.5  ins. 
lead  R.H.  P2  is  right  handed.  What  is  the  lead  of  P2? 
Which  way  must  F  turn  as  seen  from  above  (right  handed 
or  left  handed')  ?  (Two  possible  solutions.) 


J 


PROB.  94 


I — I 


96.  Screw  S  has  10  threads  per  inch  (single)  right  handed 
and  is  fixed.  Nut  A  may  slide  but  cannot  turn.  How  many 
(single)  threads  per  inch  has  screw  Si  if  46  turns  of  the  hand- 
wheel  in  the  direction  shown  lower  A  0.66  ins.?  Are  threads 
on  Si  right  handed  or  left  handed?  If  the  hand-wheel  had  a 
rim-radius  of  7  ins.  and  Si  had  8  threads  per  inch  (single)  right 
handed  what  force  would  be  necessary  at  the  rim  to  raise  a 
weight  (W)  of  16,800  Ibs.?  Allow  60  per  cent  lost  in  friction. 


PROB.  95 


PROBLEMS 


353 


96.  The  hub  H  of  the  200-tooth 
gear  forms  the  nut  for  the  screw  S. 
The  graduated  wheel  is  fast  to  the 
shaft  with  the  two  pinions.  How 
far  will  S  move  along  its  axis  when 
W  is  turned  through  the  angle  rep- 
resented by  one  division  ?  In  which 
direction  will  S  move  if  W  turns 
with  the  arrow? 


97.  If  the  mechanism  of  Prob. 
96  were  changed  as  shown  in  this 
figure,  how  far  would  S  move  and 
in  which  direction  if  W  turns  with 
the  arrow,  one  division? 


40  Th.  per  tacit 
R.M.  StctgZe 


PROB.  96 


M          1^ 
120  r,   m  ^ 

t3r'2T. 


PEOB.  97 


98.  In  this  differential  screw,  A  and  C  are 
broad-faced  pinions  which  are  fast  to  each  other 
and  are  turned  by  the  crank  shown.  H  is  a 
fixed  nut,  E  is  a  left  handed  screw  having  |  in. 
lead  and  F  is  a  right  handed  screw  having  f  in. 
lead.  How  far  does  E  move  for  24  turns  of  the 
crank  and  in  which  direction  relative  to  the 
crank  rotation? 


99.  In  a  screw  cutting  train  (Figs.  240a,  240b),  the  lead  screw  has  a  lead  of  £  in. 
left  handed.  Gear  A  has  30  teeth,  while  gear  B  has  40  teeth.  Find  suitable  num- 
bers of  teeth  for  the  change  gears  C  and  D,  using  no  gear  of  less  than  16  teeth,  if  the 


354 


PROBLEMS 


lathe  is  to  cut  threads  from  4  to  20  per  inch,  also  to  cut  1 1 1  threads  per  inch.  Arrange 
results  in  the  form  of  a  table,  using  the  least  number  of  gears  and  having  as  few 
teeth  as  possible.  Give  the  number  of  different  change  gears  that  must  be  fur- 
nished with  the  lathe  and  also  the  total  number  of  teeth  that  must  be  cut  to  make 
the  set  of  change  gears.  Should  N  and  M  both  be  used  in  cutting  a  right  handed 
thread,  or  only  M  ? 


100.  In  a  worm  and  wheel  let  the  worm  be  triple-threaded 
and  the  diameter  of  the  drum  be  14  ins.  How  many  teeth 
must  the  wheel  have  if  30  turns  of  the  worm  are  to  move  W 
20  ins.  ?  If  R  =  16|  ins.,  what  must  F  be,  if  W  equals  8800 
Ibs.  actually  lifted,  40  per  cent  being  the  loss  due  to  friction? 
If  the  handle  is  "pushed"  to  raise  the  weight,  is  the  worm 
right  handed  or  left-handed?  (Take  TT  =  -^.) 


PROB.  100 


101.  Worm  A  is  double-threaded  and  its  worm  wheel 
has  36  teeth.  Worm  B  has  a  lead  of  f  in.  The  pitch 
diameter  of  the  drum  for  the  weight  is  one  foot.  The 
force,  F,  at  the  end  of  a  16-in.  handle  on  B  is  20  Ibs.  and 
W  is  25,344  Ibs.  If  60  per  cent  is  lost  in  friction  what  is 
the  diameter  of  worm  wheel  C?  (Take  TT  =  -5Vl.) 


PKOB.  101 


B 


102.   F  is  a  double-threaded  right  handed  worm.     A  is  a  worm  wheel  having  32 
teeth.     On  the  same  shaft  with  A  is  a  gear  (7,  17-inch  pitch  diameter  in  mesh  with 
gear  D,  4-in.  pitch  diameter.     On  the  shaft  with  D 
is  the  left-handed  worm  E  having  a  lead  of  2  ins., 
in  mesh  with  worm  wheel  B,  10.83  ins.  pitch  diam. 
Disk  H  is  fast  to  the  shaft  of  worm  F,  and  B  is 
loose  on  this  shaft. 

1°  How  many  turns  of  handle  before  H  and  B 
will  be  in  the  same  position  relative  to  one  another  ? 

2°  What  changes  in  these  results  would  occur  if 
worm  E  were  right-handed  instead  of  left-handed? 

3°  If  a  drum  20  ins.  in  diameter  were  attached 
to  B  and  a  weight  of  2000  Ibs.  suspended  from  it, 
how  large  a  force  at  the  handle  would  be  necessary  to  raise  the  weight? 
friction. 


PROB.  102 


Neglect 


PROBLEMS 


355 


103.  P  is  threaded  through  the  worm  wheel  but  can- 
not turn.  Lead  of  thread  on  P  is  f  in.  right  handed. 
Worm  is  double-threaded  and  right-handed.  How  many 
turns  of  B,  and  which  way  (right-handed  or  left-handed) 
to  raise  weight!  in.  ?  What  weight  can  be  raised  by  a 
force  of  50  Ibs.  applied  at  Ft  (30  per  cent  friction  loss.) 


104.  This  plate  cam  is  made  up  of  arcs  of  two  circles 
and  their  common  tangents  and  turns  about  the  fixed 
center  A .  Plot  a  curve  showing  the  motion  of  the  follower 
for  every  15  degrees  movement  of  the  cam  for  |  turn. 
Scale  of  plot  is  to  be  as  follows:  Abscissa?  =  angles  turned 
by  cam,  f  in.  =  15  degrees.  Ordinates  =  full  size  dis- 
placements of  the  follower. 

Make,  on  the  same  plot,  a  curve  that  would  show 
the  displacements  of  the  follower,  if  its  motion  had  been 
harmonic. 


PROB.  104 


106.  The  plate  cam  with  axis  at  A  consists  of  the  arcs  of 
three  circles  with  centers  and  radii  as  shown.  Cam  turns  uni- 
formly left-handed.  Draw  a  diagram  which  shall  show  the 
motion  of  the  follower,  ordinates  to  be  distance  moved  by  the 
follower  (full  size),  and  abscissae  to  represent  angular  motion  of 
the  cam  (I  in.  =  30  degrees).  Take  points  every  30  degrees 
with  an  extra  point  at  225  degrees. 


PROB.  105 


356 


PROBLEMS 


106.  Find  the  outline  of  a  plate  cam  which,  by  turning  about 
the  center  C  as  shown  by  the  arrow,  shall  cause  a  point  A  to 
move  along  the  path  A-B  at  a  uniform  speed  as  follows :  T6o  the 
distance  A-B  in  |  turn  of  the  cam.  Still  f  turn.  Remaining 
part  of  the  distance  in  £  turn.  Return  to  A  at  once  over  the 
same  path  as  previously  traversed.  Still  J  turn.  (Take  10  in- 
tervals in  the  distance  A-B.)  Cam  to  be  full  size. 


107.  Starting  from  the  position  shown,  the 
slide  is  to  drop  2  ins.  with  harmonic  motion 
during  f  of  a  turn,  to  rise  at  once  1  inch,  to 
remain  still  |  of  a  turn,  to  drop  2  ins.  with 
uniformly  accelerated  and  uniformly  retarded 
motion  in  \  turn,  and  then  to  rise  3  ins.  at 
once.  Find  the  cam  outline  if  the  end  A  of  the 
lever  is  in  contact  with  the  cam,  the  latter  to 
turn  in  the  direction  shown.  (Assume  that  A 
is  kept  in  contact  with  the  cam  by  some  ex- 
ternal force.) 


PROS.  106 


PROB.  107 


108.  Find  the  outline  of  a  plate  cam  turning 
uniformly  right-handed  to  give  block  A  the  following 
motion:  remain  still  ^  turn,  rise  1  in.  with  harmonic 
motion  in  \  turn,  still  |  turn,  drop  If  ins.  at  once, 
rise  f  in.  uniformly  in  £  turn.  Cam  is  to  drive  roller 
B  1  in.  in  diameter. 


Fixed 


PROBLEMS 


357 


109.  Piece  A  carries  a  pin  which  projects  into 
the  slot  on  the  horizontal  piece  B.  Find  outline 
of  a  plate  cam  turning  uniformly  right-handed  to 
act  at  D  and  give  A  the  following  motion:  Still 
for  I  turn  of  cam;  up  1|  ins.  with  harmonic 
motion  in  i  turn;  still 
once,  and  still  J  turn. 


5  turn,  drop  1|  ins.  at 


Fixed 


PBOB.  109 


110.  A  and  B  are  two  rollers  (f-in.  diameter)  attached  to 
the  same  frame.  The  rollers  are  in  the  same  plane  and  both 
are  always  to  be  in  contact  with  a  single  plate  cam.  Find 
the  outline  of  the  cam  if  the  frame  is  to  be  raised  1  in.  with 
harmonic  motion  in  f  turn  of  the  cam.  What  will  be  the  motion 
of  the  follower  during  the  remaining  \  turn  of  cam? 


111.  Referring  to  Fig.  261,  p.  211,  a  cam  turning  uniformly  in  a  clockwise  direc- 
tion, on  the  axis  E,  is  to  give  the  following  motion  to  the  follower  £,  the  lowest 
position  of  the  flat  surface  of  $  being  2  ins.  above  E.    Up  2  ins.  with  harmonic 
motion  in  \  turn  of  the  cam,  down  1  in.  with  harmonic  motion  in  j  turn,  still  |  turn, 
down  1  in.  with  harmonic  motion  in  J  turn.    Find  the  shape  of  the  cam. 

112.  Referring  to  Fig.  262,  p.  212,  the  cam  turns  in  a  clockwise  direction  on 
axis  C,  the  pivot  P  for  the  lever  R  is  2£  ins.  to  the  right  and  3  ins.  above  C.    The 
radius  of  the  end  of  the  slider  $  is  f  in.    The  center  line  of  the  slide  $  is  3  ins.  to  the 
left  of  C  and  when  in  its  lowest  position,  its  lowest  point  is  2  ins.  above  C.     "T"  is 
1  in.,  and  a  line  parallel  to  the  top  surface  of  R  and  \  in.  below  it  will  pass  through 
the  pivot  P.    $  moves  up  3  ins.  with  uniformly  accelerated  and  uniformly  retarded 
motion  in  \  turn  of  the  cam,  still  |  turn,  down  3  ins.  with  uniformly  accelerated  and 
uniformly  retarded  motion  in  f  turn,  still  |  turn.    Find  the  shape  of  the  cam. 


358 


PROBLEMS 


113.  A  cylindrical  cam  3  ins.  outside  diameter  (turning  right  handed  as  seen  from 
the  right)  is  to  move  a  roller.  Roller  is  above  and  moves  parallel  to  the  axis  of  the 
cam.  Roller  moves  as  follows:  To  right  with  harmonic  motion  lj  ins.  in  £  turn  of 
cam.  Still  for  £  turn.  To  left  with  harmonic  motion  If  ins.  in  f  turn.  Still  for 
i  turn.  The  roller  is  to  be  f  in.  in  diameter  at  the  large  end  and  of  such  form  as  to 
give  pure  rolling  contact.  Groove  in  cam  is  to  be  f  in.  deep.  Draw  development 
for  both  top  and  bottom  of  groove  of  the  part  of  cam  which  causes  the  motion  to 
take  place. 


114.  A  and  D  are  fixed  axes.  AB  =  1 
in.,  BC  =  If  ins.,  BE  =  3£  ins.,  DC  =  1| 
ins.  Find  the  instantaneous  axis  of  BCH. 
Assuming  the  velocity  of  B  to  be  represented 
by  a  line  f  in.  long,  find  the  linear  velocity 
of  the  point  H  on  the  rod  BCH. 


116.  A  is  a  fixed  axis.  AB  =  1 
in.,  BC  =  2  ins.,  BH  =  3  ins.  Assum- 
ing the  velocity  of  B  to  be  represented 
by  a  line  1  in.  long,  find  velocity  of  C 
and  of  H . 


PROS.  115 


116.  A  is  a  fixed  axis.  AB  =  1  in., 
BC  =  2  ins.,  BH  =  3  ins.,  HE  =  2  ins. 
Assuming  velocity  of  B  =  1  in.,  find 
velocity  of  E. 


PBOB.  116 


117.  A  ladder  12  ft.  long  is  leaning  against  a  wall  and  makes  an  angle  of  60 
degrees  with  the  pavement.  The  wall  is  perpendicular  to  the  pavement.  If  the 
lower  end  of  the  ladder  slips  on  the  pavement  at  the  rate  of  1  ft.  per  second,  find 
graphically  the  velocity  of  the  top  of  the  ladder,  the  velocity  of  a  point  4  ft.  from 
the  lower  end  and  the  velocity  of  the  point  on  the  ladder  which  is  moving  the 
slowest  at  this  time. 


PROBLEMS 


359 


118.  The  velocity  of  A  is  repre- 
sented by  a  line  1^  ins.  long;  find 
velocity  of  block  P  if  the  wheel  rolls 
without  slipping. 


PROS.  118 

119.  The  diagram  represents  the  running  gear  of  a  locomotive.  The  driving 
wheels  A,  A  are  6  ft.  in  diameter,  while  the  trailing  wheel  B  is  3  ft.  in  diameter.  CD 
is  15  ins.  and  DE  is  60  ins.  The  locomotive  is  moving  forward  with  a  velocity 
represented  by  a  line  2  ins.  long.  Find  graphically  the  absolute  velocity  of  the 
crosshead  E,  and  its  velocity  relative  to  the  guides,  also  find  the  velocity  of  the  top 
of  the  trailing  wheel  B. 


PROB.  119 


120.  Velocity  of  A  is  represented 
by  a  line  \\  ins.  long.  Find  graphi- 
cally the  velocity  of  B  if  B  rolls  with- 
out slipping  on  C.  C  and  D  turn  to- 
gether and  D  rolls  without  slipping 
on  surface  E. 


Fixed. 


2  Dia, 


£•  75 


121.  A  is  a  disk  2  ins.  in  diameter  and  B 
is  a  disk  If  ins.  in  diameter,  both  turning 
on  fixed  axes.  A  drives  B  with  no  slipping. 
C  is  a  fixed  axis.  ED  =  3|  ins.,  CD  =  1£ 
ins.,  DX  =  If  ins.,  FX  =  3  ins.  If  the 
surface  speed  of  A  is  1  in.,  find  the  velocity 
of  X. 


PROS.  121 


360 


PROBLEMS 


122.  K,  G,  and  H  are  fixed  axes.  The 
wheel  is  2  ins.  in  diameter  and  rolls  on 
track  T  without  slipping.  KM  =  2}f 
ins.,  KA  =  1^  ins.,  BD  =  1$  ins.,  GF  =  1 
in.,  HE  =  1  in.,  DC  =  3|  ins.,  DF  =  2| 
ins.,  D#  =  l^f  ins.  If  the  velocity  of  A 
is  \  in.,  find  velocity  of  C. 


PROB.  122 


123.  Assuming  no  slip  be- 
tween the  cylindrical  disks,  and 
a  surface  velocity  of  50  ft.  per 
minute,  find  the  velocity  of  the 
slide  in  ft.  per  minute. 


PROB.  123 


2"Dla. 


124.  Assuming  no  slip  between  the  disks,  and 
a  surface  velocity  of  disk  A  =  1  in.,  find  the  velocity 
of  the  center  of  disk  X. 


125.  A,  B  and  C  are  fixed  cen- 
ters. Velocity  of  block  P  is  repre- 
sented by  a  line  1^  ins.  long.  Find 
velocity  of  P'. 


PROB.  125 


PROBLEMS  361 

126.  If  the  velocity  of  the  circumference  of  A  =  1  in.  find  the  velocity  of  X. 


PROS.  126 


127.  D  and  E  are  fixed  axes.  DA  =  3^  ins., 
DF  =  If  ins.,  EC  =  2f  ins.,  EG  =  1^  ins.,  GB  =  3 
ins.,  FB  =  2  ins.  If  the  velocity  of  A  is  2  ins.,  find 
velocity  of  B. 


PROB.  127 


128.  Given  velocities  of  A 
and  B  represented  by  lines  f 
and  11  ins.  long  respectively. 
Find  the  velocity  of  S. 


PROB.  128 


362 


PROBLEMS 


129.  Given  the  velocity  of  A  repre- 
sented by  a  line  1  in.  long,  find  the  velocity 
of  B. 


PROB.  129 

130.  A  2-pitch  pinion  of  14  teeth  is  turning  clockwise  at  the  rate  of  50  r.p.m.  and 
is  driving  a  rack.  The  tooth  of  the  rack  is  a  straight  line  making  an  angle  of  22| 
degrees  with  the  line  of  centers  and  the  addendum  of  the  pinion  is  equal  to  the 
module.  Find  by  graphical  method  the  rate  of  sliding  between  the  teeth  at  the 
pitch  point  and  at  the  end  of  recess.  Give  results  in  inches  per  minute,  assuming 
TT  =  3.  Scale,  one  inch  equals  300  ins.  per  minute. 


131.  In  this  linkage  let  BC  =  AB  =  CD  =  1± 
ins.,  and  AD  —  2f  ins.  Find  the  centrode  of  BC 
between  the  positions  BiCi  and  B2C2  and  then  find 
the  centrode  of  AD  if  BC  is  fixed. 


PROS.  131 


132.  AD  (fixed)  =  10  ins.,  BC  =  10|  ins., 
])C  =  4|  ins.  The  sketch  shows  a  linkage 
used  in  -the  feed  mechanism  of  a  vertical 
boring  mill.  The  crank  AB  is  adjustable 
for  varying  throws.  Determine  the  angular 
motion  of  DC  when  A  B  is  2  \  ins.  (this  being 
the  setting  for  maximum  feed). 


PROB.  132 


. .    angular  speed  CD  ,      _  _  , 
133.   Plot  a  curve  showing  ratio  an°ular  ^eed  AB  for  30-degree  i 

starting  with  AB  in  position  ABi  and  turning  uni- 
formly left-handed. 

Ordinates  =  angular  speed  ratio  (1  in.  =  unity). 

Abscissae  =  angular  positions,  of  AB   (|  in.  =  30 


AC  =  |  in.,  DC  =  U  ins.,  DB  =  2J  ins.,  AB  =  2 


PROB.  133 


ins. 


PROBLEMS 


363 


134.  AC  =  BD  =  8  ins.,  AB  =  CD  =  3  ins.  If 
AB  is  turning  uniformly  25  r.p.m.  calculate  the  max- 
imum angular  speed  of  CD  in  radians  per  minute. 
Sketch  the  linkage  in  the  position  at  which  the  max- 
imum angular  speed  of  CD  exists. 


PROB.  134 


135.  A  gas  engine  has  a  stroke  of  5  ins.  and  a  connecting  rod  10  ins.  long. 
Calculate,  and  check  by  graphical  construction,  the  speed  at  which  the  piston  is 
moving  when  it  is  at  mid-stroke,  if  the  engine  is  turning  1100  r.p.m.     Calculate  the 
acceleration  of  the  piston  when  the  crank  is  at  a  dead  point. 

136.  Disks  M  and  N  turn  about  their 
respective  centers,  C  and  B,  within  the  frame 
K.    Rod  L  slides  in  the  jblock  at  C  which  is 
fixed  to  disk  M.    The  center  line  of  the  block 
coincides  with  a  diameter  of  M.    What  is  the 


value  of  ratio 


angular  speed  disk  M 


angular  speed  disk  N 
position  shown? 


in  the 


137.  A  swinging-block  quick  return  motion  has  a  ratio tlme  °f,  cutting  str(*e  =  ?• 

time  of  return  stroke      1 

If  the  driving  crank  is  1  in.  long,  locate  the  fixed  point  of  the  swinging  link.  Draw 
the  linkage  in  some  convenient  position,  draw  the  infinite  links  which  have  been 

replaced  by  the  sliding  pair,  state  the  ratio  angular  speed  driven  crank  ^  ^ 

angular  speed  driving  crank 

position  (numerical  values  not  required).     State  maximum  (numerical)  value  of  this 
ratio. 

138.  In  the  swinging  block  mechanism,  Fig.  335,  let  the  maximum  value  of 
BA  =  8  ins.  and  the  minimum  value  =  3  ins.;  the  arm  CN  =  3  ft.;  NP  =  2  ft. 
2  ins.;  path  of  P  is  perpendicular  to  CB  and  19  ins.  above  B;  maximum  value  of 
time  of  cutting  stroke      2 

time  of  return  stroke  =  T;  angular  speed  of  «ear  "  =  30  r'P'm- 
1°  Find  position  of  axis  C. 

««IB»    «     •  •  i        P  time  of  cutting  stroke 

2°  Find  minimum  value  of  -. 7 — 7 —      .     ,     • 

time  of  return  stroke 

3°  Plot  a  curve  whose  abscissae  are  time  units  and  whose  ordinates  are  linear 
speeds  of  P  in  feet  per  minute.  Scale  of  abscissae  |  in.  =  time  occupied  by  gear  M 
in  turning  through  an  angle  of  15  degrees.  Scale  of  ordinates  1  in.  =  100  ft.  per 
minute. 

139.  In  a  Whitworth  Quick  Return  mechanism  similar  to  Fig.  340,  BA  =  5f  ins. 
BC  =  12|  ins.    Assume  that  R  is  on  a  perpendicular  to  AB  passing  through  A 
instead  of  above  it,  as  shown  in  the  figure. 

1°  Find  the  ratio  of  time  of  cutting  stroke  to  time  of  return  stroke. 

2°  Let  AN  =  15  inches;  NR  =  4  feet;  angular  speed  of  large  gear  =  1  radian 
per  unit  of  time.  Find  the  maximum  linear  speed  of  R  in  feet  per  unit  of  time. 

3°  If  the  path  of  R  is  20  ins.  above  A,  other  dimensions  remaining  the  same 
as  before,  what  difference  results  in  ratio  of  time  of  cutting  stroke  to  time  of  return 
stroke? 


364 


PROBLEMS 


140.  Draw  a  pantograph  to  connect  two  points  A  and  B,  1£  ins.  apart,  so  that 
the  motion  of  A  shall  be  to  the  motion  of  B  as  13  is  to  7.     Calculate  the  distance 
from  B  to  the  fixed  point.    The  pantograph  is  to  be  so  arranged  that  A  may  move 
at  least  5  ins.  in  either  direction  along  the  line  through  A  and  B. 

141.  Design  a  pantograph  to  reduce  the  motion  of  the  crosshead  of  an  engine 
that  has  a  stroke  of  12  ins.,  to  3  his.,  so  that  a  steam  engine  indicator  may  be  operated 
from  it.     Let  the  distance  from  the  fixed  point  to  the  point  of  connection  at  the 
crosshead  be  15  ins.     Draw  half  size. 

142.  The  three  points  A,  B,  and  C  are  to  be  connected  by  a  pantograph  so  that 
A  may  move  up  4  ins.,  B  up  3  ins.,  and  C  down  2  ins.  ,  „  . 
AC  =  6  ins.    Locate  the  fixed  point  and  the  point  B,  and          |c          gt          *| 
then  draw  a  pantograph  that  will  allow  A  to  be  moved  4 

ins.  in  every  direction. 


B 
PROS.  142 


143.  P  is  to  have  a  stroke  of  3 
ins.  and  is  to  lie  on  the  same  vertical 
line  in  its  mid-position  (shown  in 
sketch)  and  its  two  extreme  posi- 
tions.   Locate  D  and  draw  in  the 
linkage.     Then   by  producing  AB 
and  adding  only  two  links,  design  a 
pantograph  which  shall  contain  a 
point  whose  motion  is  parallel  to 
that  of  P  and  equal  to  f  of  it.    Draw 
in  the  pantograph    and 

the  point  clearly. 

144.  Stroke  of  P=  2  inches 
in  the  vertical  straight   line, 
locate  the  points  C  and  B  giv- 
ing the  link  BC  and  the  moving 
point  P.    Then  connect  this 
point  with  a  point  R,  2  ins. 
horizontally  to  the  right  of  P, 
by  a  pantograph  so  that 

motion  of  R  _  3 
motion  of  P  ~  2 

and  calculate  the  distance  from 
R  to  the  fixed  point  of  the 
pantograph. 


PROB.  143 


~dl 


'\ 


T 


PROB.  144 


145.  Referring  to  Fig.  374,  the  stroke  of  e  is  f  of  an  inch  downward  from  the 
position  shown.  A  pencil  at/ draws  a  line  along  ss  4  ins.  long,  the  highest  position 
of  /  being  2  ins.  above  line  ca.  ss  is  parallel  to  and  4|  ins.  to  the  right  of  tt.  ca  is 
perpendicular  to  tt  and  1|  ins.  above  highest  position  of  c.  Link  ab  is  If  ins.  long. 
Link  dc  swings  equal  angles  either  side  of  a  line  through  d  parallel  to  tt.  Locate  d, 
find  length  of /c,  dc,  be  and  he;  locate  a  and  h.  Solve  by  calculation  or  graphically, 
as  is  more  convenient.  Draw  and  dimension  the  mechanism. " 


PROBLEMS 


365 


///777777//7////T// 


146.   Find  W  if  there  is  a  friction  loss  of 
40  per  cent.  F=75  Ibs. 


147.  In  this  hitch,  what  force  F  is  required  to  raise  a  weight 
W  of  1400  Ibs.,  friction  being  neglected? 


148.  If  W  =  3000  Ibs.,  find  the  force  F 


149.  Two  men,  weighing  150  Ibs.  each,  stand  on  W  and  pull  just 
enough  to  sustain  the  load. 

(a)  What  pull  do  they  exert  on  the  rope? 

(6)  What  is  the  tension  on  the  support  for  the  upper  block, 
neglecting  the  weight  of  the  blocks  and  rope  itself  ? 

(c)  If  the  men  stood  on  the  ground  what  would  be  the  tension  in 
the  rope  which  supports  the  upper  block? 


366  PROBLEMS 

150.  A  differential  pulley-block  is  to  lift  1500  Ibs.  with  a  pull  of  30  Ibs.,  friction 
being  neglected.     Find  the  ratio  of  the  larger  diameter  of  the  upper  sheave  to  the 
smaller  one. 

151.  In  a  differential  pulley-block,  the  smaller  diameter  of  the  upper  sheave  is 
12  ins.    It  is  found  necessary  to  haul  over  7  ft.  of  chain  to  raise  the  weight  6  ins. 
What  is  the  other  diameter  of  the  upper  sheave?     Neglecting  friction,  what  weight 
would  be  raised  by  a  pull  of  40  Ibs. 

152.  With  a  differential  pulley-block,  if  the  diameters  of  the  sheaves  in  the  fixed 
block  are  12  ins.  and  11  ins.,  and  if  the  weight  of  the  lower  block  is  20  Ibs.,  what  net 
weight  can  be  raised  by  a  pull  of   120  Ibs.  on  the  chain,  allowing  a  loss  of  30  per 
cent  in  friction?     How  much  chain  must  be  overhauled  to  lift  the  weight  one  foot? 


INDEX 


Acceleration,  angular,  6. 

linear,  5. 

normal,  6. 

tangential,  6. 
Acting  flank,  94. 
Action,  angle  of,  96. 

arc  of,  96. 
Addendum,  94. 

circle,  94. 

limits  of,  on  involute  gears,  112. 
Aggregate  combinations,  306. 

motion  by  linkwork,  306. 
Anchor  escapement,  331. 
Angle  of  action,  96. 

of  approach,  96. 

of  recess,  96. 

pressure,  97. 
Angular  acceleration,  6. 
Angular  speed,  5. 
Annular  gear,  86. 

involute,  117. 

cycloidal,  128. 
Approach,  angle  of,  96. 

arc  of,  96. 
Arc  of  action,  96. 

of  approach,  96. 

of  recess,  96. 
Automobile  differential,  173,  177. 

transmission,  160,  172. 
Axial  pitch,  149. 
Axis,  instantaneous,  228. 

Backlash,  95. 

Bands,  3. 

Barrel,  54. 

Bearings,  10. 

Bell  crank  lever,  16. 

Belt,  approximate  length  of,  28. 

calculation  of  power,  25. 

connecting  non-parallel  shafts,  37. 

crossed,  23. 

double,  24. 


Belt  drives,  examples  of,  40-49. 

exact  length  of,  29. 

open,  23. 

quarter-turn,  38. 

single,  24. 

tension  in,  25. 
Belting,  power  of,  24. 
Belts,  21-50. 

kinds  of,  23. 

Bevel  epicyclic  train,  176-179. 
Bevel  gears,  86,  144-148. 

twisted,  86. 
Block  chains,  57. 
Bush,  10. 

Cam  and  slotted  sliding  bar,  327. 
Cam,  cylindrical,  197,  213-219. 

definition  of,  197. 

diagram,  198-200. 

plate,  197,  200-212. 

plate,  with  flat  follower,  210-212. 

positive  motion  plate,  209. 
Cams,  combination  of,  219-220. 
Centrode,  229. 

of  a  rolling  body,  229. 
Chain,  kinematic,  244. 
Chains,  21,  55-62. 

block,  57. 

calculation  for  length  of,  59. 

conveyor,  55. 

hoisting,  55. 

Morse  rocker-joint,  62. 

power  transmision,  57 

Renold,  60. 

roller,  59. 

silent,  60. 

Chronometer  escapement,  334. 
Chuck,  elliptic,  283. 
Circular  pitch,  95. 
Clearance,  94. 
Clearing  curve,  104. 
Click,  311. 


367 


368 


INDEX 


Clockwork,  157. 

Closed  pair,  9. 

Collar,  11. 

Combination,  elementary,  1. 

Components,  225. 

Composition  of  velocities,  226. 

Compound  screws,  188. 

Cones,  rolling  without  slip,  67-72. 

Conic  four-bar  linkage,  286. 

Conjugate  curves,  99. 

construction  of,  99. 
Connecting  rod,  221. 
Contact,  path  of,  97. 

pure  rolling,  63. 
Continuous  motion,  4. 
Conveyor  chains,  55. 
Cords,  21,  50. 

small,  53. 

Cotton  card  train,  161. 
Counter  mechanism,  322,  325. 
Coupling,  Oldham's,  284. 
Crank,  16,  221. 

and  rocker,  250. 

pin,  221. 

Crown  gears,  86,  147. 
Crown-wheel  escapement,  330. 
Crowning  of  pulleys,  49. 
Cycloidal  gears,  122-138. 

low  numbered  teeth,  135. 
Cylinder  and  sphere,  rolling,  73. 
Cylindrical  cam,  197,  213-219. 

multiple  turn,  215-218. 
Cylinders,  rolling  without  slip,  63-66. 

Dead-beat  escapement,  332. 

Dead  points,  249. 

Diametral  pitch,  95. 

Differential,  automobile,  173,  177. 

Differential  screws,  188. 

Direction,  3. 

Disk  and  roller,  74. 

Drag  link,  252. 

Driven  wheel,  153. 

Driver,  2. 

Driving  wheel,  153. 

Drum,  54. 

Eccentric,  275. 

rod,  275. 
Eccentricity,  275. 


Effective  lever  arms,  20. 
Effective  pull,  '25. 
Elementary  combination,  1. 
Elements,  expansion  of,  274. 

pairs  of,  8. 

Ellipses,  rolling,  81,  82. 
Elliptic  chuck,  283. 

trammel,  281. 
Engine,  oscillating,  270. 
Epicyclic  train,  166-179. 
Epitrochoid,  104. 
Escapement,  311. 

anchor,  331. 

chronometer,  334. 

crown-wheel,  330. 

dead-beat,  332. 

Graham  cylinder,  333. 
Escapements,  330. 
Expansion  of  elements,  274. 

Face,  of  gear,  94. 

of  tooth,  94. 

width  of,  94. 
Feather,  11. 

Ferguson's  paradox,  171. 
Flank,  acting,  94. 

of  tooth,  94. 
Flexible  connectors,  21. 
Follower,  2. 
Foot-pound,  24. 
Four-bar  linkage,  221-245. 

angular  speed  ratio,  246. 

diagrams  for  change  in  speed  ratio, 
248. 

non-parallel  crank,  254. 

parallel  crank,  254. 

relative  motion  of  links,  246. 

sliding  slot,  273. 

swinging  block,  267. 

turning  block,  271. 
Frame,  2. 

Frequency  of  contact,  156. 
Friction  catch,  318. 
Friction  gearing,  74-76. 

grooved,  76. 

Gear  teeth,  involute,  106-122. 

law  governing,  98. 

rate  of  sliding  of,  241. 
Gear  train  design,  162-165. 


INDEX 


369 


Gears,  annular,  86. 

automobile  differential,  173,  177. 

bevel,  144-148. 

classified,  86. 

crown,  86,  147. 

cycloidal,  122-138. 

cycloidal,  low  numbers  of  teeth,  135. 

drives,  86. 

epicyclic,  166-179. 

face  of,  94. 

helical,  86,  149-152. 

helical,  speed  ratio,  152. 

herringbone,  86, 139. 

hyperboloidal,  86. 

interchangeable,  119. 

involute,  standard  proportions  of,  122. 

mitre,  86. 

pin,  139-143. 

reduction,  162. 

screw,  148. 

separation  of,  119. 

skew,  86. 

skew  bevel,  147. 

sliding  eliminated,  139. 

stepped,  138. 

twisted,  138. 

twisted  bevel,  147. 
Geneva  stop,  325. 
Graham  cylinder  escapement,  333. 
Gudgeon,  10. 
Guide  pulleys,  40. 
Guides,  10. 

Harmonic  motion,  7. 
Helical  gears,  86,  149-152. 

speed  ratio,  152. 
Helix,  the,  148. 
Herringbone  gears,  86,  139. 
Higher  pair,  9. 
Hookes'  joint,  287-290. 
Horsepower,  24. 

Horsepower  of  belt,  calculation  of,  25. 
Hoisting  chains,  55. 
Hoisting  machine  tram,  162. 
Hunting  cog,  156. 
Hyperbolas,  rolling  of,  83. 
Hyperboloidal  gears,  86. 


Instantaneous  axis,  228. 

of  a  rolling  body,  229. 
Instantaneous  center,  229. 
Interchangeable,  involute  gears,  119. 
Intermittent  motion,  4. 

from  continuous  motion,  324. 

from  reciprocating  motion,  311. 
Inversion  of  pairs,  10. 
Involute,  applied  to  gear  teeth,  106. 
Involute  gears,  separation  of,  119. 
Involute,  of  a  circle,  105. 
Isosceles  linkage,  278-284. 

Journal,  10. 

Keys,  11, 
Keyway,  11. 
Kinematic  chain,  244. 

Lead,  149. 

Lever,  double  rocking,  254. 

nipping,  319. 
Lever  arms,  effective,  20. 
Levers,  16-20. 

bell  crank,  16. 

kind  of,  16. 

motion  from,  17. 
Line  of  centers,  221. 
Line  of  connection,  22. 
Linear  acceleration,  5. 

speed,  4. 

Link,  3,  222,  244. 
Linkage,  244. 

angular  speed  ratio,  246. 

conic,  286. 

four-bar,  221,  245. 

isosceles,  278-284. 

relative  motion  of  links,  246. 

sliding  block,  259. 

sliding  slot,  273. 

turning  block,  271. 
Links,  relative  motion  of,  224. 
Linkwork,  slow  motion  by,  257. 
Linkwork  with  a  sliding  pair,  258. 
Locking  devices,  328. 
Logarithmic  spirals,  rolling,  78-81. 
Lower  pah*,  9. 


Idle  wheel,  154. 
Inclined  plane,  180,  181. 


Machine,  1. 
Mashed  wheels,  321. 


370 


INDEX 


Mechanism,  1. 

constructive,  2. 

pure,  2. 

Mitre  gears,  86. 
Module,  95. 

Morse  rocker-joint  chain,  62. 
Motion,  3. 

continuous,  4. 

cycle  of,  4. 

from  levers,  17. 

graphical  representation  of,  224. 

harmonic,  7. 

intermittent,  4. 

kinds  of,  7. 

modification  of,  8. 

parallelogram  of,  225. 

parallelepiped  of,  226. 

period  of,  4. 

reciprocating,  4. 

resultant,  225. 
Multiple  turn  cylindrical  cam,  215-218. 

Neck,  10. 

Nipping  lever,  319. 

Non-cylindrical  surfaces,  rolling  of,  77- 

85. 

Normal  acceleration,  6. 
Normal  pitch,  109,  149. 

relation  to  circular  pitch,  110. 
Nut,  10. 

Obliquity  of  action,  97. 
Oldham's  coupling,  284. 
Oscillating  engine,  270. 
Oscillation,  4. 

Pau*s,  lower,  9. 

higher,  9. 

incomplete,  9. 

inversion  of,  10. 

of  elements,  8. 
Pantograph,  298. 
Parabolas,  rolling,  83. 
Parallel  motion  by  cords,  305. 

by  means  of  four-bar  linkage,  303. 
Parallelogram  of  motion,  225. 
Parallelepiped  of  motion,  226. 
Path,  3. 
Path  of  contact,  97. 


Pawl,  312. 

double  acting,  315. 

reversible,  313. 
Peaucellier's   straight   line   mechanism, 

291. 

Pedestal,  10. 
Periphery-speed,  14. 
Pillow-block,  10. 
Pin  gears,  139-143. 
Pitch,  axial,  149. 

circles,  94. 

circular,  95. 

diametral,  95. 

normal,  109,  149. 

number,  95. 

point,  94. 

surface,  22. 
Pivot,  10. 

Plane,  inclined,  180,  181. 
Planer  drive,  158. 
Plate  cam,  197,  200-212. 

with  flat  follower,  210. 
Plumber-block,  10. 
Positive  motion  plate  cam,  209. 
Power  transmission  chains,  57. 
Power,  unit  of,  24. 
Pressure  angle,  97 
Pulley,  21. 
Pulley-block,  triplex,  175. 

Weston  differential,  310. 
Pulley-blocks,  307. 
Pulleys,  crowning  of,  49. 

guide,  40. 

stepped,  30 

stepped,  equal,  34. 

tight  and  loose,  50. 

Quarter-turn  belt,  38. 
Quick  return,  swinging  block,  267-270. 
Whitworth,  271. 

Rack  and  pinion,  86. 

involute,  116. 
Radian,  12. 
Ratchet  wheel,  311. 
Recess,  angle  of,  96. 

arc  of,  96. 

Reciprocating  motion,  4. 
Reduction  gear,  162. 
Renold  silent  chain,  60. 


INDEX 


371 


Resolution  and  composition  of  velocities, 

typical  problems,  230-243. 
Resolution  of  velocities,  226. 
Resultant,  225. 
Resultant  motion,  225. 
Revolution,  4. 
Revolving  bodies,  12. 

angular  speed  of,  12. 

linear  speed  of,  12. 

Robert's  straight-line  mechanism,  302. 
Rocker,  16. 
Roller  chains,  59. 

Rolling  bodies,  velocities  of  point  on,  230. 
Rolling  contact,  pure,  63. 
Root,  circles,  94. 

distance,  94. 
Rope  driving,  51-55. 

systems  of,  51. 

grooves  for,  53. 
Ropes  and  cords,  50. 
Ropes,  wire,  54. 

wire,  grooves  for,  55. 
Rotation,  4. 

axis  of,  4. 

direction  of,  4. 

plane  of,  4. 

Scott  Russell,  straight-line  mechanism, 

292-294. 
Screw,  10. 

cutting,  190-193. 

gears,  148. 

rotation  caused  by  axial  pressure,  190. 

threads,  181. 
Screws,  differential,  188. 

speed  ratio,  187. 

Shafts  connected  by  belt,  directional  re- 
lation, 22. 

speed  ratio,  22. 
Silent  chains,  60. 
Silent  feed,  320. 
Skew  bevel,  147. 

gears,  86. 
Slides,  10. 

Sliding-block  linkage,  259-266. 
Sliding  slot  linkage,  273. 
Speed,  angular,  5. 

cones,  36. 

linear,  4. 

periphery,  14. 


Speed,    relation    between    forces    and, 
15. 

surface,  14. 

uniform,  5. 

variable,  5. 
Spindle,  10. 
Spline,  11. 

Sprockets,  diameters  of,  60. 
Spur  gears,  86. 

twisted,  86. 
Star  wheel,  327. 
Step,  10. 
Stepped  pulleys,  30. 

for  crowned  belt,  31. 

for  open  belt,  33. 
Stepped  wheels,  138. 
Straight-line  mechanism,  291. 

Peaucellier's,  291. 

Robert's,  302. 

Scott  Russell,  292-294. 

Tchebicheff's,  303. 

Watt's,  295-298. 
Sun  and  planet  wheel,  170. 
Surface  speed,  14. 
Swash-plate,  285. 
Swinging-block  linkage,  267. 

Tangential  acceleration,  6. 
Tchebicheff's  straight  line  mechanism, 

303. 
Threads,  left-hand,  186. 

per  inch,  186. 

right-hand,  186. 

screw,  181-187. 
Tight  and  loose  pulleys,  50. 
Tram,  automobile  transmission,  160. 

bevel  epicyclic,  176-179. 

cotton  card,  161. 

design  of,  162-165. 

epicyclic,  166-179. 

hoisting,  162. 

value,  154. 

Trains  of  wheels,  153. 
Trammel,  elliptic,  281. 
Turn,  4,  12. 

Turning  block  linkage,  271. 
Transmission,  automobile,  160,  172. 

modes  of,  2. 

Triple  pulley  block,  175. 
Twisted  bevel  gears,  86,  147. 


372 


INDEX 


Twisted  spur  gears,  138. 
Uniform  speed,  5. 
Universal  joint,  288. 

Velocities,  composition  and  resolution  of, 

226. 

graphical  representation  of,  224. 
of  rigidly  connected  points,  226. 
typical  problems,  230-243. 


Velocity,  5. 
Vibration,  4. 

Watt's  straight-line  mechanism,  295-298. 

Wedge,  180-181. 

Whitworth  quick  return,  271. 

Wiper,  197. 

Wire  ropes,  54. 

Worm  and  wheel,  86,  193-196. 


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